# 2. Differential Equations of First Order and First Degree – Differential Equations

## Differential Equations of First Order and First Degree

##### 2.1 FIRST ORDER AND FIRST DEGREE DIFFERENTIAL EQUATIONS

The simplest type of the ordinary differential equations is that of the first order and first degree and is of the form

Solving for we can write Eq. (2.1) as

and if we write

then Eq. (2.2) can be put in the form

where M and N are functions of x and y. In order to study the methods of solution, we classify the first order and first degree equations as follows:

1. Variables separable
2. Reducible to variables separable
3. Homogeneous equation
4. Non-homogeneous but reducible to homogeneous type
5. Exact equation
6. Inexact equation but rendered exact using integrating factors
7. First order linear equation.
8. Reducible to linear equation
1. Bernoulli's equation
2. Others

Now we take up the above types of equations one by one and discuss the methods of solution.

#### 2.1.1 Variable Separable Equation

If the first order and first degree differential equation is of the form f1(x)g2 (y)dx + f2(x)g1(y)dy = 0 then it can be written as M(x)dx + N(y)dy = 0 where M(x) = (f1/f2)(x), is a pure function of x; and N(y) = (g/1/g2)(y) a pure function of y, so that the equation can be readily integrated and the solutions obtained.

Example 2.1.1   Solve e−4y log x dx + x cos3y dy = 0.

Solution   If we multiply the equation by it becomes . Integrating, we get

where c is an arbitrary constant.

Example 2.1.2   Solve

Solution   Separating the variables, we have x(2log x + 1)dx − (sin y + y cos y)dy = 0. Integrating, we get x2 log x = y sin y + c.

Example 2.1.3   Solve

Solution   Separating the variables,

Integrating, we get

where c = 2c1.

Example 2.1.4   Solve

Solution   Put x + y + 2 = u

Integrating, we get

Example 2.1.5   Solve (1+ x2)dy = (1+ y2)dx.

Solution

Separating the variables

Integrating (2.5) we get

Example 2.1.6   Solve

Solution

Separating the variables we get

Integrating (2.9) we get

Example 2.1.7   Solve

Solution

Separating the variables we get

Integrating (2.12) we get

Dropping logs we have

Example 2.1.8   Solve

Solution

The given equation becomes

Separating the variables

This can be written in the form

Integrating we get

Dropping logs we get

Example 2.1.9   Solve

Solution

Separating the variables

Integrating (2.22) we have

Dropping logs

##### EXERCISE 2.1

Solve:

1. Ans: x + a = c(y + b)

2. Ans:

3. Ans:

4. Ans: y sin−1 x = c

5. Ans: sin−1 x + sin−1 y = c

6. Ans:

7. ydxxdy + 3x2y2ex2dx = 0.

Ans:(x/y) + ex3 = c

8. (y2x2)y′ + 2xy = 0.

Ans: x4 + 2e2x = c + 4ey

9. Ans:

10. Ans:

11. Ans:

12. Ans: tan[(x + y) 2] = x + c

13. Ans: y sin y = x2 log x + c

14. Ans: (x − 1)(y − 1) = cxy

15. Ans: x + y + xy = 1

16. Ans: x(y + 4) = c

17. Ans:

18. Ans: (y − 3)(3x +1) = cx

19. (ey + 1)cos x dx + ey sin x dy = 0.

Ans: (1 + ey)sin x = c

20. Ans:

21. Ans:

22. cos(x + y)dy = dx.

Ans:

23. Ans:

24. Ans:

#### 2.1.2 Homogeneous Equations

A polynomial f (x, y) is called a homogeneous function of degree n if f (tx, ty) = tn f (x, y) for some tR or equivalently, f (x, y) = xn f (1, y/x).

Example 2.1.10 f (x, y) = 2x + 3y is a homogeneous function of degree 1 since

Example 2.1.11 g (x, y) = 3x2 − 4xy + 7y2 is a homogeneous function of degree 2 since

A first order and first degree differential equation

is called a homogeneous type if f(x,y) is a rational function which is homogeneous and of degree zero. That is, f(x,y) is the quotient of two polynomials each of the same degree n. If we put y=vx (or x=vy) Eq. (2.26) transforms to

where the variables are separable. Separating the variables and integrating, we get the general solution as

where v = y/x (or v = x/y) and c is an arbitrary constant.

We extend the meaning of ‘homogeneity’ to include differential equations with functions of the following type since they are solvable by the above method.

Example 2.1.12

Example 2.1.13

Example 2.1.14

Example 2.1.15   Solve

Solution   Put y = vx

The equation transforms to

Separating the variables

Integrating, log x + log v + v = constant

Dropping logarithms we may put the solution as

where c is an arbitrary constant.

Example 2.1.16   Solve .

Solution   Put

The equation becomes

Separating the variables

Integrating

Dropping logarithms, replacing v by and simplifying we get

where c is an arbitrary constant.

Example 2.1.17   Solve

Solution   Put     x = vy

The equation transforms to

Separating the variables we have

where c is an arbitrary constant.

Example 2.1.18   Solve .

Solution

Eq. (2.28) can be written as

This is a homogeneous equation. Putting or y = vx in (2.29) we get

Separating the variables, we have

Integrating we have

Example 2.1.19   Solve

Solution

Eq. (2.33) can be written as

Putting x = vy we get

Separating the variables and integrating

##### EXERCISE 2.2

Solve:

1. Ans:

2. Ans:

3. Ans:

4. Ans: cx=(x2y2)

5. x2ydx=(x3+y3)dy.

Ans:

6. (x2y2)dx=xydy.

Ans: x2(x2−2y2) = c

7. Ans:

8. Ans:

9. Ans:

10. Ans: x10=cy6(x2y2)

11. Ans: xecos(y/x) = c

12. Ans:

13. Ans:

14. Ans:

15. Ans: xy(xy)=c

16. Ans:

17. Ans:

18. Ans:

19. (xy log y + y log x)dx + x(log y − log x)dy = 0.

Ans: (xy)log x + y log y = cx + y

20. x2dy = (x2 + xy + y2)dx.

Ans:

#### 2.1.3 Non-homogeneous Equations

The first order and first degree differential equation of the type

where a, b, c, l, m, n are given real constants is called a non-homogeneous differential equation.

If c = n = 0 then Eq. (2.34) reduces to a homogeneous equation whose solution we have discussed above. The new method of solution (the method of proportions) used for solving non-homogeneous equations (case (b)) below may be used for solving the homogeneous equation of this type. But other forms under homogeneous type have to be solved only by putting y = vx or x = vy. We assume that at least one of c and n is not zero. In general, the following two cases are discussed.

Case (A) In this case, Eq. (2.34) can be solved by putting u = ax + by or u = lx + my and separating the variables.

Case (B) In this case, Eq. (2.34) is solved by shifting the origin to a point (h, k) by the substitution x = X + h, y = Y + k and choosing (h, k) such that ah + bk + c = 0; lh + mk + n = 0 which is possible since Indeed, we have Eq. (2.34) now reduces to the homogeneous type which can be solved by putting Y = vX or X = vY and separating the variables. This method is lengthy and tedious.

By dividing the problem into various cases, we can apply simpler and easier methods to solve the problem. Finally, the technique described below, based on the application of the property of proportion makes solution of the problem simpler and easier in the above two cases, as well. In this section, C is used to denote an arbitrary constant instead of c.

Case (1) (say). Eq. (2.34) reduces to whose general solution is y = λx + c.

Case (2) (i) b = l = 0 or (ii) a = m = 0.

In either case, the variables are separable and hence Eq. (2.34) is integrable.

Case (3) c ≠ 0 or n ≠ 0 or both c ≠ 0 and n ≠ 0.

(i) a = b = 0 (c ≠ 0) (ii) l = m = 0 (n ≠ 0)

In case (3)(i) Eq. (2.34) reduces to

which can be solved by putting u = lx + my and in case (3)(ii), Eq. (2.34) reduces to

which can be solved by putting u = ax + by

#### Alternative method (Method of proportions)

By the property of proportions, we can write Eq. (2.35) in case (3) as

using (l,m) as multipliers, each ratio

Integrating, the last two ratios give

or, dropping logs we get

(lx + my + mc/l)c = Cely

Now, by the property of proportions, we can write Eq. (2.36) in case (3)(ii) as

Using (a,b) as multipliers, each ratio

Integrating, the last two ratios give

or dropping logs we get

(ax + by + an/b)n = Cebx

Case (4) b + l = 0.

Separating the variables and grouping the terms, we can write Eq. (2.34) as

(ax + c)dx + bd(xy) − (my + n)dy = 0

l = −b and ydx + xdy = d(xy)

Integrating, we obtain the general solution as

ax2 + 2bxymy2 + 2cx − 2ny + C = 0

This is an exact equation. We will study exact equations in detail is Section 2.1.4.

Case (5) a = m, b = l ≠ 0.

Eq. (2.34) becomes

This falls under case (A) if a = b and case (B) if ab.

But it can be solved simply by the method of proportions as follows. Eq. (2.37) can be written as

Integrating we have

or, dropping logs we get

Case (6) a = −m, b = −l.

Eq. (2.34) can be written as

Cross-multiplying we get

[(b + a)(y + x) + cn]d(y + x) = [(ba)(yx)+ c + n]d(yx)

Integrating we have

[(b + a)(y + x)2 + (ab)(yx)2 + 2(cn)(y + x) −(c + n)(yx) = K

Note 2.1.20 This problem is also solvable by grouping terms as in case 4.

Finally, we consider case (A) and case (B) mentioned at the beginning.

Case (7) b + l ≠ 0.

(i) and (ii) respectively.

(i) Alternative method (Method of proportions)

Equation (2.34) can be written as

Using (λ, −1) and (l, m) as multipliers, we obtain from the last equation that each ratio

Integrating, we get

(ii) Alternative method (Method of multipliers) m ≠ 0; Δ = (bl)2 + 4am ≠ 0

Equation (2.34) can be written as

Using (1,λ) as multipliers, each ratio

Now, we determine λ such that

which is a quadratic in λ and has two real distinct roots λ1, λ2 if Δ = (bl)2 + 4am > 0. More specifically

Also,

∴ each ratio

Integrating and dropping logs, the general solution of equation (2.34) in this case may be written as

Note 2.1.21 If ∆ < 0 then the roots λ1, λ2 are complex conjugate numbers of the form α ± . The solution (2.47) is valid in this case also. In physical applications we have to separate the real and imaginary parts of (2.47) and take either of them as the solution of the problem. If ∆ = 0 or m = 0 the method is not applicable. Then we have to apply the method of case (A), case (B) or case (2)(ii) as suitable.

Example 2.1.22   Solve .

Solution   Assume that 3xy + 2 ≠ 0. Here

 a = 3, b = −1, c = 2 l = 6, m = −2, n = 4

Since the differential equation reduces to (case 1) whose general solution is .

Example 2.1.23   Solve .

Solution   Here

Since b = l = 0, the variables are separable (case 2(i)). Separating the variables, we have

(2x + 3)dx − (4y − 5)dy = 0

Integrating, we get the general solution as

x2 − 2y2 + 3x + 5y + C = 0

Example 2.1.24   Solve .

Solution   Here

Since a = m = 0, the variables are separable (case (2)(ii)). Separating the variables, we have

Integrating we have

Dropping logs and simplifying we get

(4y − 3) = C(2x + 5)2

Example 2.1.25   Solve .

Solution   Here

Putting u = 3x + y we can separate the variables and solve.

#### Alternative method (Method of proportions)

The given equation can be written as

Using (3, 1) as multipliers, each ratio

Integrating we get

Example 2.1.26   Solve .

Solution   Here

Putting u = x − 2y we can separate the variables and solve.

#### Alternative method (Method of proportions)

The given equation can be written as

Using (1, –2) as multipliers, each ratio

Integrating

Dropping logs, the solution can be expressed as

(2x − 4y + 5)3ex2 = C

Example 2.1.27   Solve

Solution   Here

Separating the variables and grouping the terms we can write the differential equation as

on multiplying by 2 and noting that ydx + xdy = d (xy).

Integrating, we have the general solution as

x2 + 2xyy2 − 4x + 8y + C = 0

Example 2.1.28   Solve .

Solution   Here .

Put x = X + h, y = Y + k. The differential equation reduces to provided (h, k) are chosen so that 2h + 5k + 1 = 0 and 5h + 2k − 1 = 0.

By the rule of cross-multiplication

Separating the variables we have

Integrating we get

Multiplying by 2 and dropping logs we have

But so that

∴ The general solution is

(3y − 3x + 2)7 = C(y + x)3.

#### Alternative method (Method of proportions)

Here a = m = 2, b = l = 5.         (case (5))

The given differential equation can be written as

Integrating

Simplifying we can write the general solution as

(3y − 3x + 2)7 = C(y + x)3

Example 2.1.29   Solve

Solution   Here

Put x = X + h, y = Y + k and choose h, k so that

6h + 5k − 3 = 0 and 4h + k + 2 = 0

Now the differential equation transforms to which is a homogeneous equation.

Putting Y = vX; the equation becomes

Putting into partial fractions

Integrating and dropping logs

Let

Absorbing the constants in the arbitrary constants C we can write the solution as

(2y − 6x − 9)7 = C(7y + 14x + 1)2

#### Alternative method (Method of proportions)

By the property of proportions and by using (1, λ) as multipliers, we can write the differential equation as

Choose λ so that .

For λ = −3 and 2 we get from the last term

Integrating, we have

Dropping logs, after simplification,

(2y − 6x − 9)7 = C(7y + 14x + 1)2

Example 2.1.30   Solve .

Solution   By the property of proportions, we have

Choose λ such that

Each ratio

Integrating we get

where .

Example 2.1.31   Solve .

Solution   By the property of proportions and by using (1, λ) as multipliers we have

Substituting the two values and λ2 = −2 for λ we get

Integrating and dropping logs we obtain

(2y + 3x + 1)2(2y − 4x − 13)5 = C

Example 2.1.32   Solve .

Solution   Here the method of proportions is not applicable since we obtain roots λ1 = λ2 = −1 for λ. We apply the usual method of case (B).

Putting x = X + h, y = Y + k and determining h, k from

Then the given equation reduces to .

Put .

Separating the variables and integrating, after multiplying by 2,

we get

.

The required solution is

##### EXERCISE 2.3

Solve the following differential equations:

1. Ans: y = 3x + C

2. Ans: 3x2 − 2y2 + 4x + 6y + C = 0

3. Ans: (y + 3)2 = C(2x − 7)

4. Ans: (25x − 35y + 2)9 = C e5y

5. Ans: (10x − 25y − 52)11 = C e−5x

6. Ans: y2 + 2xy − x2 − 2y − 4x + C = 0

7. Ans: y2 + xyx2x − 3y + C = 0

8. Ans: 2x2 + 2xyy2 + 12x + 6y + C = 0

9. Ans: x2 + 3xy + 2y2 + xy + C =0

10. Ans: x2 + 2xyy2 − 4y + C = 0

11. Ans: x2 − 6xy + 9y2 + 8x + 14y + C =0

12. Ans: (yx)3 = C (y + x − 2)

13. Ans: (yx + 1)a+b(y + x − 1)ab = C

14. Ans: y = x + log(x + y) + c

15. (2x2 + 3y2 − 7)xdx + (3x2 + 2y2 − 8)ydy = 0.

[Hint: Put x2 = X,y2 = Y and group the terms:
(2X − 7)dX + 3d(XY) + (2Y − 8)dY = 0]

Ans: X2 + 3XY + Y2 − 7X − 8Y + C = 0

or x4 + 3x2y2 + y2 − 7x2 − 8y2 + C = 0

16. Ans: u2uvv2 + u + C = 0 or e2x + ex = C + ex+y + e2y

17. Ans: 2x2 + 2hxy + by2 + 2gx + 2fy + C = 0

#### 2.1.4 Exact Equations

Definition 2.1.33 A differential equation which is obtained from its primitive by mere differentiation without any further operation is called an exact equation.

The equations xdy + ydx = 0, xdx + ydy = 0 and are exact since these can be written as and respectively.

The following theorem gives a criterion for exact equations.

Theorem 2.1.34 If M(x, y) and N(x, y) are real-valued functions having continuous first partial derivatives on some rectangle , then a necessary and sufficient condition for the equation

to be exact in R is that

Proof

(i) The condition is necessary: Suppose Eq. (2.48) is exact. Then there exists a function u(x, y) such that

Differentiating M and N partially with respect to y and x, respectively

From Eqs. (2.52) and (2.53) it follows that

(ii) The condition is sufficient: Assume that Eq. (2.49) holds.

where the integral is evaluated partially with respect to x treating y as constant so that .

Hence, the equation becomes

which proves that the equation is exact.

#### Procedure for solving an exact equation

1. Test if
2. Compute

(or)

3. Compute

(or)

4. The general solution is

(or)

Note 2.1.35 In most cases, collecting and grouping terms which are exact differentials will not only prove the exactness of the equation but also yield the general solution, on integration, as the illustrative examples will show.

Note 2.1.36 The following formula

though in many cases, gives the correct solution, fails in some cases and is therefore not advisable.

(Example 2.1.37 shows the failure of this method to give the correct solution.)

Example 2.1.37   Solve .

Solution   Here,

The equation is exact, since

Hence, the general solution is

Note 2.1.38 If we compute the solution using Eq. (2.57) we get

in which the term is missing.

Example 2.1.39   Solve .

Solution   In the standard form, the equation is

(3x − 2y + 5)dx + (−2x + 4y + 1)dy = 0

M = 3x − 2y + 5; N = −2x + 4y + 1

My = −2 = Nx = −2

The equation is exact since

The general solution is

#### Alternative method: Grouping the terms

This shows that the equation is exact. Integrating it, we get

Example 2.1.40   Solve .

Solution   Writing the equation in the standard form

(3x − 2y + 5)dx + (−2x + y2 − 2y)dy = 0

M = 3x2 − 2xy ∢ 5; N = −x2y2 + 2y

My = −2x = Nx = −2x

The equation is exact,

The general solution is

#### Alternative method: Grouping the terms

The equation is exact; integrating and multiplying by 3, we get the general solution as

3x3 − 3x2y + 3y2 − 15x + c = 0.

Example 2.1.41   Solve

Solution   Here,

The equation is exact.

The general solution is .

#### Alternative method: Grouping the terms

The equation is exact.

The general solution is .

Example 2.1.42   Solve (y2 − 2xy)dx = (x2 − 2xy)dy.      [JNTU 1995]

Solution

M = y2 − 2xy; N = 2xyx2

My = 2y − 2x = Nx = 2y − 2x

The equation is exact.

The general solution is xy2x2 y = c.

#### Alternative method: Grouping the terms

(The equation is exact.)

The general solution is xy2x2 y = c.

Example 2.1.43   Solve .

Solution

(dividing the numerator and denominator by x2)

From the above results, we observe that the equation is exact. Its general solutions is

Example 2.1.44   Solve (ey + 1)cos x dx + ey sin x dy = 0.

Solution

(The equation is exact.)

Integrating, we get the general solution as

(ey + 1)sinx = c.
##### EXERCISE 2.4

Solve:

1. (2x + 3y + 4)dx + (3x − 6y + 2)dy = 0.

Ans: x2 + 3xy − 3y2 + 4x + 2y + c = 0

2. (3x2 − 9x2y2 + 2xy)dx + (6y2 − 6x3y2 + x2)dy = 0.

Ans: x3 + 2y3 − 3x3y2 + x2y = c

3. ex(sin x + cos x)sec y dx + exsin x sec y tan y dy = 0.

Ans: ex sin x sec y = c

4. (cos x cos y − cot x)dx = sin x sin y dy.

Ans: sin x cos y = log |c sin x|

5. Ans: xy + y log x + x sin y = c

6. 4y sin2xdx + (2y + 3 − 4cos2 x)dy = 0.

Ans: y2 + y − 2y cos2x = c

7. Ans:

8. (x2ay)dx = (axy2)dy.

Ans: x3 + y3 = 3axy + c

9. xy2 = cey2

Ans:

10. Ans: xy + y sin x + x sin y = c

11. Ans:

12. Ans:

#### Integrating factor (I.F.)

If the differential equation Mdx + Ndy = 0 becomes exact when we multiply it by a function µ (x, y) then µ (x, y) is called an integrating factor of the equation.

Consider the equation ydxxdy = 0

Here

The equation is not exact. If we multiply it by we get

Now so that

So, the equation becomes exact. is an integrating factor. We can easily check that are also integrating factors for the equation.

Integrating factor can be found by inspection, after grouping of terms. Table 2.1 gives the list of integrating factors.

Table 2.1

Example 2.1.45   Solve the following differential equations after finding the integrating factor in each case:

Solution   The suitable integrating factor in each case is shown against each differential equation. After multiplying by the integrating factor, we can derive the general solution as follows.

1.

2. ydx + xdy + xy(ydxxdy) = 0

##### EXERCISE 2.5

Find the integrating factors by inspection and solve the following:

1. xdy − (y − 3x2)dx = 0.

Ans:

2. yex + exdy = 2xy2dx.

Ans:

3. xdy + 2ydx = 2y2xdy.

Ans:

4. (x2 + y2)(xdyydx) = (x2y2)(xdx + ydy).

Ans: (x + y) = c(xy)(x2 + y2)

5. 2xydyy2dx + x3exdx = 0.

Ans: y2 + x(x − 1)ex + cx = 0

Given below are the rules for finding integrating factors of Mdx + Ndy = 0 depending on the nature of functions M and N.

Rule 1 If Mdx + Ndy = 0 is a homogeneous differential equation and Mx + Ny ≠ 0, then is an integrating factor of the equation.

Note 2.1.46 Recall that we have already solved homogeneous differential equations reducing them to variables separable form by putting y = vx or x = vy.

Rule 1 Provides an alternative method for its solution.

Note 2.1.47 If Mx + Ny = 0 then and the equation becomes ydxxdy = 0 whose general solution is

Rule 2 If Mdx + Ndy = 0 is of the form

f1(xy)ydx + f2(xy)xdy = 0

and MxNy ≠ 0 then is an integrating factor of the equation.

Rule 3 If purely a function of x alone, then is an integrating factor of the equation Mdx + Ndy = 0.

Note that the linear equation where P, Q are functions of x alone may be written in the form Mdx + Ndy = 0 where M = PyQ and N = 1 so that

purely a function of x and the integrating factor in this case is , as we will see in linear equations discussed below.

Rule 4 If , purely a function of y alone then is an integrating factor of the equation Mdx + Ndy = 0.

Rule 5 If Mdx + Ndy = 0 is expressible in the form

xayb(mydx + nxdy) + xcyd(pydx + qxdy) = 0

where a,b,c,d,m,n,p,q are all constants such that mqnp ≠ 0 then xh yk is an integrating factor of the equation for some suitable constants h, k to be determined from the two equations and

Example 2.1.48   Solve (x2 + y2)dxxydy = 0 by finding an integrating factor.

Solution   This is a homogeneous differential equation.

Here,

Since , the equation is not exact

Mx + Ny = (x2 + y2)x − (xy)y = x3 ≠ 0

Multiplying by the integrating factor,

∴ The general solution is

Example 2.1.49   Find an integrating factor and solve

Solution   Putting the equation in the standard form (xy + x2 + y2) dxx2dy = 0

Since , the equation is not exact

The given equation is a homogeneous differential equation.

Multiplying the equation by the integrating factor

Example 2.1.50   Solve (xy + 1) ydx + (−xy + 1) xdy = 0.

Solution   This equation is of the form

f1(xy)ydx + f2(xy)xdy = 0

By Rule 2, the integrating factor is if MxNy ≠ 0.

MxNy = (xy + 1)yx −(−xy + 1)yx = 2x2y2 ≠ 0

Multiplying by the integrating factor the equation can be written as

Example 2.1.51   Solve (x2y2 + xy + 1) ydx + (x2y2xy + 1) xdy = 0.

Solution   The equation is of the form

f1(xy)ydx + f2(xy)xdy = 0

Integrating factor

Multiplying the equation by and omitting the constant,

The general solution is .

Example 2.1.52   Solve (xy)dxdy = 0.

Solution

(Constant can be considered as a function of x.)

By Rule 3, the integrating factor is .

Multiplying by the integrating factor = ex the equation can be written as

Example 2.1.53   Solve (3xy − 2y2)dx + (x2 − 2xy)dy = 0.

Solution

By Rule 3, the integrating factor is .

Multiplying by integrating factor = x, the equation becomes

(3x2y − 2xy2)dx + (x3 − 2x2y)dy = 0
d(x3y) = d(x2y2) ⇒ x3y = x2y2 + c

is the general solution of the differential equation.

Example 2.1.54   Solve (xy3 + y)dx + 2(x2y2 + x + y)dy = 0.

Solution

a function of y alone.

By Rule 4, the integrating factor is .

Multiplying by integrating factor, the equation can be written as

General solution is

Example 2.1.55   Solve (y + xy2)dxxdy = 0.

Solution

By Rule 4, the integrating factor is

a pure function of y.

Multiplying the equation by we can write the equation as , whose general solution is

Example 2.1.56   Solve xy3 (ydx + 2xdy) + (3ydx + 5xdy) = 0.

Solution   The given differential equation is

Here    M = xy4 + 3y     N = 2x2y3 + 5x

My = 4xy3 + 3     Nx = 4xy3 + 5

Since MyNx, the equation is not exact. We can easily verify that Rules 1–4 are not applicable here. So, we try to find an I.F. of the form xhyk, by applying Rule 5.

Comparing Eq. (2.58) with the standard form

We have

The constants h, k are determined from

An integrating factor is x2 y4. Multiplying Eq. (2.58) by this integrating factor, we have

which, on regrouping and expressing as exact differentials, yields

which is the required solution.

Example 2.1.57   Solve x(3ydx + 2xdy) + 8y4 (ydx + xdy) = 0.

Solution   The given differential equation is

We observe that an integrating for this equation is of the form xh yk for some h, k Multiplying Eq. (2.66) by xh yk, it can be written in the form

where       M = 3xh+1yk+1 + 8xhyk+5,

N = 2xh+2yk + 8xh+1yk+4

Nx = 2(h + 2)xh+1yy + 8(h + 1)xhyk+4

Exactness condition is

Now the left-hand side expression of Eq. (2.66) is exact

which is the required solution.

Example 2.1.58   Solve 2x2(ydx + xdy) + y(ydxxdy) = 0.

Solution   The given differential equation is

 Here M = 2x2y + y2 N = 2x3 − xy ∴ My = 2x2 + 2y Nx = 6x2 − 2 y

Since MyNx, the equation is not exact.

To find an integrating factor of the form xhyk, we multiply Eq. (2.70) by xhyk so that new value of

M = 2xh+2yk+1 + xhyk+2 and

N = 2xh+3ykxh+1yk+1

My = 2(k + 1)xh+2yk + (k + 2)xhyk+1

Nx = 2(h + 3)xh+2yk − (h + 1)xhyk+1

Equating My = Nx, we have

Integrating, we get

which is the required solution.

Example 2.1.59   Solve xy(ydx + xdy) + x2y2(2ydxxdy) = 0.

Solution   Multiplying the equation by an integrating factor xh yk, we obtain

Equating the coefficients of like powers

Now Eq. (2.71) becomes

Integrating, we get

the required general solution.

##### EXERCISE 2.6

Solve the following equations by finding integrating factors:

1. (x2y − 2xy2)dx − (x3−3x2y)dy = 0.

(Madras 1975, Karnataka 1971, Calcutta Hon. 1975)

Ans: (x/y) + log(y3/x2) = c

2. (x2 + y2)dx − 2xydy = 0.

Ans: x2y2 = cx

3. x2ydx − (x3 + y3)dy = 0.

Ans:

4. y(xy + 2x2y2)dx + x(xyx2y2)dy = 0.

(Rajasthan 1969, Kanpur 1974, Karnataka 1971, Punjab 1971)

Ans:

5. (x3y3 + x2y2 + xy + 1)ydx + (x3y3x2y2xy + 1)xdy = 0.

(Kanpur 1980, Gorakhpur 1972)

Ans:

6. (xy sin xy + cos xy)ydx + (xy sin xy − cos xy)xdy = 0.

(Gorakhpur 1974, Kanpur 1977, Lucknow 1975, Rajasthan 1976, Marathawada 1974)

Ans: (x/y)sec xy = c

7. (x2 + y2 + 2x)dx + 2ydy = 0.

(Srivenkateswara 1984, Calicut 1983)

Ans: ex(x3 + y3) = c

8. (y + log x)dxxdy = 0.

Ans: ex + y + log x + 1 = 0

9. (3x2y4 + 2xy)dx + (2x3y3x2)dy = 0.

(Calcutta Hon. 1952, 1954; Utkal 1980)

Ans:

10. (xy3 + y)dx + 2(x2y2 + x + y4)dy = 0.

Ans: 3x2y4 + 6xy2 + 2y6 = c

11. x(4ydx + 2xdy) + y3(3ydx + 5xdy) = 0.

Ans: x4y2 + x3y3 = c

12. xy3(ydx + 2xdy) + (3ydx + 5xdy) = 0.

Ans:

#### 2.1.6 Linear Equations

A differential equation of the form

where p0, p1 and p2 are continuous functions of x in some interval I, is called a first order linear differential equation in y. Dividing Eq. (2.73) by p0 we can write it as

which is taken as the standard form of the first order linear equation. Here the dependent variable y and its derivative occur separately and to a first degree. For example,

If Q(x) ≢ 0, then Eq. (2.73) is called a non-homogeneous linear equation. If Q(x) ≡ 0, then it is called a homogeneous or reduced linear equation.

Writing Eq. (2.73) in the form

and comparing it with

we have M = PyQ,N = 1.

Since

a function of x alone is an integrating factor of Eq. (2.74). Multiplying Eq. (2.74) by the integrating factor , we have

Separating the variables and integrating, we get the general solution of Eq. (2.74) as

where c is an arbitrary constant.

#### Method of solving a linear equation

1. Write the equation in the standard form

2. Find the integrating factor and multiply the equation by the integrating factor.

3. Write the general solution of the differential equation as y(integrating factor) = c + ∫Q(integrating factor)dx, after evaluating the integral.

Note 2.1.60 The words homogeneous and non-homogeneous used here are not to be confused with similar words used earlier.

Note 2.1.61 Sometimes a linear equation may be written in the form

so that Eq. (2.79) can be written as d(P0y = Qdx which, on integration, yields the general solution

Example 2.1.62

Solution   This can be written as

Separating the variables and integrating

(1 + x2)y = logsinx + c

where c is an arbitrary constant.

Note 2.1.63 An equation which is not linear in y may be linear in the variable x.

Example 2.1.64   Solve y2dx + (xy − 1)dy = 0.

Solution   Rewriting this equation as

which is linear in x.

Integrating factor =

x(integrating factor) =

Integrating factor dy

Example 2.1.65   Solve .

Solution   The equation is linear in y

Here

Multiplying the equation by x, we have

Integrating    xy = c + x (ex − cos x)

= c + x(ex − cosx) − ex + cosx
xy = c + (x − 1)(ex − cosx).

Example 2.1.66

Solution   Writing the equation in the standard form

Here

This is a linear equation in y

The general solution is

[Hint: put logx = t,

Example 2.1.67   Solve .

Solution   This equation can be written as

The equation is linear in x

The general solution is

Example 2.1.68   Solve

Solution   Writing the equation as

The equation is linear in x

The general solution is

Put tan−1 y = t,

Example 2.1.69   Solve

Solution   Writing the equation as

Here P(x) = sec2 x,     Q(x) = tan x sec2 x. This is linear in y. Also,

P(x)dx = ∫sec2xdx = tanx

Integrating factor = etanx

The general solution is

yetanx = c + ∫tanxsec2xetanxdx

Put tan x = t,       sec2 xdx = dt

= c + ∫tetdt
= c + (t − 1)et

Integrating by parts

y = ce−tanx + (tanx − 1).

Example 2.1.70   Solve xy′ + y + 4 = 0.          [JNTU 2001]

Solution   Writing the equation in the form

we note that this is a linear equation in y with

But, as it is, the equation can be written as

d(xy) + 4dx = 0

which, on integration, yields the general solution

xy + 4x + c = 0

Example 2.1.71   Solve

Solution   This is a linear equation in y with

But, as it is, the equation can be written as

The general solution is

(x2 − 1) y = x + c

Example 2.1.72   Solve

Solution   This is a linear equation in y with

But, as it is, it can be written as

d(xy) = logxdx
∫logxdx = xlogxx

Integrating, the general solution is

xy = c + (x − 1) log x

Example 2.1.73   Solve .

Solution   The equation can be put in the standard form

This is linear in y with

The general solution is

Example 2.1.74   Solve

Solution   Divide by x2

Example 2.1.74   Solve

Solution   Divide by x2

Example 2.1.74   Solve

Solution   Divide by x2

Example 2.1.75   Solve

Solution   Divide by (x + 1)n+1

Example 2.1.76   Solve

Solution

Eq. (2.82) can be written as

Here P = tan x; Q = sec x

Multiplying (2.83) by sec x, we have

Example 2.1.77   Solve

Solution

Multiplying Eq. (2.85) by , we get

or on integration, we obtain

Example 2.1.78   Solve

Solution

Eq. (2.89) can be written as

Eq. (2.90) is linear in x; Here

Multiplying (2.90) by we get

whose solution is

##### EXERCISE 2.7

Solve the following:

1. Ans:

2. Ans: xy sec x = tan x + C

3. Ans: y = Ce−2x + ex sin 2x

4. y′ + ytan x = sin 2x, y(0) = 1

[Hint: y sec x = sin2 x + c]

or y = cos x − cos3 x + ccos x

y(0) = 1 ⇒ 1 = c.

Ans: y = 2cos x − cos3 x

5. dx = (x + y + 1)dy.

Ans: x + y + 2 = cey

6. y′ + ycot x = 2x cosec x.

Ans: y = (x2 + c)cosec x

7. Ans: 16x2y + x4(1 − 4log x) = c

8. Ans: 4y sec x = 2x + sin 2x + c

9. Ans:

10. Ans:

11. Ans: x sec y = tan y + c

12. yeydx = (y3 + 2xeydy).

Ans:

13. Ans:

14. Ans:

15. Ans:

16. [Hint: ]

on multiplication by sec2 x

⇒ ∫ d (y tan x) = ∫ xexdx.

Ans: y tan x = c + (x − 1)ex

17. Ans: y = tan x − 1 + ce−tan x

18. Ans: yx sec x = tan x + c

19. Ans: y log x = c + (log x)2

20. Ans:

21. Ans: sin y = (1 + x)(ex + c)

22. Ans: 2x = ey(1 + x2)

23. Ans:

24. (1 + y + x2y)dx + (x + x2)dy = 0.

Ans: xy = c − tan−1 x

#### 2.1.7 Bernoulli's Equation

An equation of the form

is called Bernoulli's equation.

Bernoulli's equation is reducible to the linear form by the substitution z = y1−a.

Multiplying Eq. (2.91) by (1 − a)ya, it becomes

which is linear in z. Its general solution is

Note 2.1.79 There are also equations that are not of the Bernoulli type shown in Eq. (2.91). These equations are reducible to linear form by appropriate substitution (See Example 2.1.92).

Example 2.1.80   Solve .

Solution   This is Bernoulli's type equation.

P = −tanx, Q = −secx, a = 2.

Multiplying the equation by , we have

(This is a linear equation.)

Integrating factor

ePdx = etanx = elogsecx = secx

The general solution is

Example 2.1.81   Solve y(2xy + ex)dx = exdy.

Solution

This is Bernoulli's equation.

P = −1, Q = 2xe−x, a = 2

Multiplying the equation by , we have

This is a linear equation in

Integrating factor = e∫1dx = ex

The general solution is

Example 2.1.82   Solve .

Solution

This is linear in .

Multiplying by , we have

The general solution is

Example 2.1.83   Solve .

Solution   This not strictly in the form of Bernoulli's equation (2.91). But we can write it as

which is a Bernoulli type Eq. (2.91).

The general solution is

##### EXERCISE 2.8

Solve the following:

1. Ans: (5 + cx2)x3y5 = 2

2. Ans:

3. Ans: (log y)−1 = 1 + cx

4. Ans: y2x2 = cx − 1

5. x(xy)dy + y2dx = 0.

Ans:

6. Ans: cos y = cos x(sin x + c)

7. Ans: 2 (tan x + sec x) = y(2sin x + 1)

8. y (2xy + ex)dx = exdy.

Ans: ex = y(cx2)

9. Ans: sec y = (c + sin x)cos x

10. Ans: sin y = (1 + x)(ex + c)

11. Ans:

12. Ans:

13. Ans:

14. y(2xy + ex)dxexdy = 0.

Ans: ex + yx2 + cy = 0

15. Ans:

16. Ans:

#### Newton's Law of Cooling

Physical experiments show that the rate of change of temperature T with respect to time t,dT/dt, of a body is proportional to the difference between the temperature of the body (T) and that of the surrounding medium (T0).

This principle is known as Newton's Law of Cooling and is expressed through the following first order and first degree differential equation:

Separating the variables

Integrating, we get

If Ti is the initial temperature of the body when t = 0, from Eq. (2.97)

Eliminating c between Eqs. (2.97) and (2.98), we get

#### Method of solving the problem of Newton's Law of Cooling

1. Identify T0, the temperature of the surrounding medium. Then the general solution is given by Eq. (2.99).
2. Use two given conditions and find the constant of integration c and the proportionality constant k.
3. Substitute c and k obtained from step 2 in Eq. (2.99). We can determine (i) the value of T for a given time t or (ii) the value of t for a given temperature T from Eq. (2.99).

#### Law of Natural Growth or Decay

If the rate of change of a quantity y at any time t is proportional to y, then

If k is the constant of proportionality, then the required differential equation is

where k is a real constant.

For growth, k > 0, the differential equation is

For decay, the differential equation is

Example 2.2.1   The temperature of a body initially at 80°C reduces to 60°C in 12 min. If the temperature of the surrounding air is 30°C, find the temperature of the body after 24 min.

Solution   Let T be the temperature of the body at time t.

By Newton's Law of Cooling

Temperature of the surrounding medium T0 = 30

Initial temperature T = Tt = 80 when t = 0

Example 2.2.2   A body is heated to 105°C and placed in air at 15°C. After 1 hr its temperature is 60°C. How much additional time is required for it to cool to

Solution   Let T be the temperature of the body at time t.

By Newton's Law of Cooling

Temperature of the surrounding medium T0 = 15

Initial temperature (t = 0)

Additional time required = 2 hr – 1 hr = 1 hr.

Example 2.2.3   The number N of bacteria in a culture grew at a rate proportional to N. The value of N was initially 100 and increased to 332 in 1 hr. What was the value of N after hrs? [JNTU 1996S, 2003]

Solution   Here N is a natural number and is therefore discrete. But in view of its largeness N will be treated as a continuous variable which is a differentiable function of time t. The differential equation to be solved is

Example 2.2.4   A radioactive substance disintegrates at a rate proportional to its mass. When the mass is 10 mg, the rate of disintegration is 0.051 mg per day. How long will it take for the mass of 10 mg to reduce to half.

[JNTU 1995]

Solution   The governing differential equation is

where m is the mass of the substance.

Separating variables and integrating

When mass m = 10mg

negative sign is to be taken since is decreasing rate.

Eq. (2.114) becomes

We have to find t when m = 5 mg

##### EXERCISE 2.9
1. A body initially at 80°C cools down to 60ºC in 20 min. The temperature of the air is 40°C. Find the temperature of the body after 40 min.

Ans: 50°C

2. The air temperature is 20°C. A body cools from 140°C to 80°C in 20 min. How much time will it take to reach a temperature of 35°C?

Ans: 60 min

3. If the temperature of the air is 20°C and the temperature of a body drops from 100°C to 75°C in 10 min. what will be its temperature after 30 min? When will the temperature be 30°C?

Ans: 46°C, 55.5 min

4. Uranium disintegrates at a rate proportional to the amount present at any instant. If m1 and m2 gms of uranium are present at time t1 and t2, respectively. Show that the half-life of uranium is [(t2t1)log 2]log(m1 / m2).

5. The rate at which bacteria multiply is proportional to the instantaneous number present. If the original number doubles in 2 hrs, in how many hours will it triple? [JNTU 1987, 2000]

Ans:

6. The rate of decay of radium varies as its mass at a given time. Given that half-life of radium is 1600 yrs, find out the percentage of the mass of radium it will disintegrate in 200 yrs.

Ans: 8.3% approximately

#### Orthogonal trajectories of a family of curves

Definition 2.2.5 A curve which cuts every member of a given family of curves at a right angle is called an orthogonal trajectory of the given family of curves.

A family of curves which are orthogonal to themselves are called self-orthogonal.

1. In the electrical field, the paths along which the current flows are the orthogonal trajectories of the equipotential curves.
2. In fluid mechanics, the stream lines and the equipotential lines are orthogonal trajectories of one another.
3. In thermodynamics, the lines of heat flow are perpendicular to isothermal curves.

#### (1) Cartesian coordinates

represent the equation of a given family of curves with single parameter ‘c’.

Differentiating Eq. (2.118) with respect to ‘x’ and eliminating ‘c’ from the equation thus obtained and Eq. (2.118) we obtain a differential equation of the form

for the given family of curves.

Suppose there passes a curve of the given family and a member of the orthogonal trajectories through a point P (x, y). Let m1 be the slope of the curve of the given family and m2 the slope of an orthogonal trajectory. Since the curves cut at right angles, we have

so that

Therefore, if we replace in Eq. (2.119) by we get the differential equation for the orthogonal trajectories as

Integrating Eq. (2.120), we obtain the equation for the orthogonal trajectories.

Example 2.2.6   Find the orthogonal trajectories of the rectangular hyperbolas x2y2 = c where c is a parameter.

Solution   Equation of the given curves

Differentiating Eq. (2.121) with respect to x, we get the differential equation of the given curves as

Replacing in Eq. (2.122) by the differential equation for the orthogonal trajectories is obtained as

Separating the variables, we get

Integrating, the equation for orthogonal trajectories is

Example 2.2.7   Find the orthogonal trajectories of the family of parabolas through the origin and focii on the y-axis.

Solution   Equation of the family of parabolas

Differentiating Eq. (2.125) with respect to x,

Eliminating ‘a’ between Eqs. (2.125) and (2.126), the differential equation for the given curves is

Replacing in Eq. (2.127) by , the differential equation for orthogonal trajectories is

Separating the variables and integrating, we get the equation for orthogonal trajectories as

2y2 + x2 = c

where ‘c’ is a constant.

Example 2.2.8   Find the orthogonal trajectories of the family of semicubical parabolas ay2 = x3 where ‘a’ is a parameter.

Solution   The equation of the given curves is

Differentiating Eq. (2.129) with respect to ‘x’, we get

Eliminating ‘a’ between Eqs. (2.129) and (2.130), the differential equation is obtained as

Replacing in Eq. (2.131) by the differential equation for the orthogonal trajectories is obtained as

Separating the variables and integrating, we obtain the equation for orthogonal trajectories as

3y2 − 2x2 = c

where ‘c’ is an arbitrary constant.

Example 2.2.9   Show that the system of confocal ellipses is self-orthogonal.

Solution   The equation of the system of ellipses

Differentiating Eq. (2.133) with respect to ‘x

Eliminating λ from Eq. (2.133) using Eq. (2.135)

It is clear that if we replace y′ by , the same equation is obtained.

Example 2.2.10   Show that the system of confocal and coaxial parabolas y2 = 4a(x + a) is self-orthogonal.

Solution   The equation of the given parabolas

Differentiating Eq. (2.136) with respect to ‘x’, we have

Eliminating ‘a’ from Eqs. (2.136) and (2.137), we get

Replacing y1 by in the above differential equation for the given curves, we obtain the same equation for the orthogonal trajectories.

#### (2) Polar coordinates

be the equation of the given system of curves in polar coordinates where ‘a’ is a parameter.

Differentiating Eq. (2.139) with respect to ‘θ’ we get another equation. Between these two equations, we eliminate the parameter ‘a’ and obtain the differential equation for the given system of curves as

If we replace in Eq. (2.140) by (which amounts to interchanging the role of the polar subnormal and the polar sub-tangent), we get the differential equation for the orthogonal trajectories as

Solving this differential equation, we get the equation for the orthogonal trajectories.

The following solved examples illustrate the procedure.

Example 2.2.11   Find the orthogonal trajectories of the family of cardioids r = a(1 − cosθ) where ‘a’ is a parameter.

Solution   The equation of the given cardioids is

Differentiating Eq. (2.142) with respect to ‘θ’, we have

Eliminating ‘a’ between Eqs. (2.142) and (2.143), we get

The differential equation for the given curves

Replacing by we obtain from Eq. (2.144) the differential equation for orthogonal trajectories as

Integrating we get      (2.145)

which is the equation for the orthogonal trajectories.

Example 2.2.12   Find the equation of the system of orthogonal trajectories of the family of curves rnsin = an where an is a parameter.

Solution   The given curves are

By logarithmic differentiation of Eq. (2.147) with respect to ‘θ

which is the differential equation for the given curves.

Replacing by in Eq. (2.148), we get the differential equation for the orthogonal system as

By integrating, we obtain the equation for orthogonal trajectories as

##### EXERCISE 2.10

Find the orthogonal trajectories of the family of curves given in Table 2.2.

Table 2.2