2
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:
 Variables separable
 Reducible to variables separable
 Homogeneous equation
 Nonhomogeneous but reducible to homogeneous type
 Exact equation
 Inexact equation but rendered exact using integrating factors
 First order linear equation.
 Reducible to linear equation
 Bernoulli's equation
 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 f_{1}(x)g_{2} (y)dx + f_{2}(x)g_{1}(y)dy = 0 then it can be written as M(x)dx + N(y)dy = 0 where M(x) = (f_{1}/f_{2})(x), is a pure function of x; and N(y) = (g/_{1}/g_{2})(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 x^{2} log x = y sin y + c.
Example 2.1.3 Solve
Solution Separating the variables,
Integrating, we get
where c = 2c_{1}.
Example 2.1.4 Solve
Solution Put x + y + 2 = u
Integrating, we get
Example 2.1.5 Solve (1+ x^{2})dy = (1+ y^{2})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:

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

Ans:

Ans:

Ans: y sin^{−1} x = c

Ans: sin^{−1} x + sin^{−1} y = c

Ans:
 ydx − xdy + 3x^{2}y^{2}e^{x2}dx = 0.
 (y^{2} − x^{2})y′ + 2xy = 0.
Ans: x^{4} + 2e^{2x} = c + 4e^{y}

Ans:

Ans:

Ans:

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

Ans: y sin y = x^{2} log x + c

Ans: (x − 1)(y − 1) = cxy

Ans: x + y + xy = 1

Ans: x(y + 4) = c

Ans: (y − 3)(3x +1) = cx
 (e^{y} + 1)cos x dx + e^{y} sin x dy = 0.
Ans: (1 + e^{y})sin x = c

Ans:

Ans:
 cos(x + y)dy = dx.
Ans:

Ans:

Ans:
2.1.2 Homogeneous Equations
A polynomial f (x, y) is called a homogeneous function of degree n if f (tx, ty) = t^{n} f (x, y) for some t ∈ R or equivalently, f (x, y) = x^{n} 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) = 3x^{2} − 4xy + 7y^{2} 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
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.
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:

Ans:

Ans:

Ans:

Ans: cx=(x^{2}−y^{2})
x^{2}ydx=(x^{3}+y^{3})dy.
Ans:
 (x^{2}−y^{2})dx=xydy.
Ans: x^{2}(x^{2}−2y^{2}) = c

Ans:

Ans:

Ans:

Ans: x^{10}=cy^{6}(x^{2}−y^{2})

Ans: xe^{cos(y/x)} = c

Ans:

Ans:

Ans:

Ans: xy(x−y)=c

Ans:

Ans:

Ans:
 (x − y log y + y log x)dx + x(log y − log x)dy = 0.
Ans: (x − y)log x + y log y = cx + y
x^{2}dy = (x^{2} + xy + y^{2})dx.
Ans:
2.1.3 Nonhomogeneous 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 nonhomogeneous 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 nonhomogeneous 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
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
Case (4) b + l = 0.
Separating the variables and grouping the terms, we can write Eq. (2.34) as
∵ l = −b and ydx + xdy = d(xy)
Integrating, we obtain the general solution as
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 a ≠ b.
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
Crossmultiplying we get
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 = (b − l)^{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 λ_{1}, λ_{2} if Δ = (b − l)^{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 λ_{1}, λ_{2} are complex conjugate numbers of the form α ± iβ. 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 3x − y + 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
Integrating, we get the general solution as
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
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
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
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 crossmultiplication
Separating the variables we have
Integrating we get
Multiplying by 2 and dropping logs we have
But so that
∴ The general solution is
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
Example 2.1.29 Solve
Solution Here
Put x = X + h, y = Y + k and choose h, k so that
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
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,
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 λ_{2} = −2 for λ we get
Integrating and dropping logs we obtain
Example 2.1.32 Solve .
Solution Here the method of proportions is not applicable since we obtain roots λ_{1} = λ_{2} = −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:

Ans: y = 3x + C

Ans: 3x^{2} − 2y^{2} + 4x + 6y + C = 0

Ans: (y + 3)^{2} = C(2x − 7)

Ans: (25x − 35y + 2)^{9} = C e^{5y}

Ans: (10x − 25y − 52)^{11} = C e^{−5x}

Ans: y^{2} + 2xy − x^{2} − 2y − 4x + C = 0

Ans: y^{2} + xy − x^{2} − x − 3y + C = 0

Ans: 2x^{2} + 2xy − y^{2} + 12x + 6y + C = 0

Ans: x^{2} + 3xy + 2y^{2} + x − y + C =0

Ans: x^{2} + 2xy −y^{2} − 4y + C = 0

Ans: x^{2} − 6xy + 9y^{2} + 8x + 14y + C =0

Ans: (y − x)^{3} = C (y + x − 2)

Ans: (y − x + 1)^{a+b}(y + x − 1)^{a−b} = C

Ans: y = x + log(x + y) + c
(2x^{2} + 3y^{2} − 7)xdx + (3x^{2} + 2y^{2} − 8)ydy = 0.
[Hint: Put x^{2} = X,y^{2} = Y and group the terms:
(2X − 7)dX + 3d(XY) + (2Y − 8)dY = 0]Ans: X^{2} + 3XY + Y^{2} − 7X − 8Y + C = 0
or x^{4} + 3x^{2}y^{2} + y^{2} − 7x^{2} − 8y^{2} + C = 0

Ans: u^{2} − uv − v^{2} + u + C = 0 or e^{2x} + e^{x} = C + e^{x+y} + e^{2y}

Ans: 2x^{2} + 2hxy + by^{2} + 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 realvalued 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
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.)
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
M_{y} = −2 = N_{x} = −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 + y^{2} − 2y)dy = 0
M = 3x^{2} − 2xy ∢ 5; N = −x^{2} − y^{2} + 2y
M_{y} = −2x = N_{x} = −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
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 (y^{2} − 2xy)dx = (x^{2} − 2xy)dy. [JNTU 1995]
Solution
M = y^{2} − 2xy; N = 2xy − x^{2}
M_{y} = 2y − 2x = N_{x} = 2y − 2x
The equation is exact.
The general solution is xy^{2} − x^{2} y = c.
Alternative method: Grouping the terms
(The equation is exact.)
The general solution is xy^{2} − x^{2} y = c.
Example 2.1.43 Solve .
Solution
(dividing the numerator and denominator by x^{2})
From the above results, we observe that the equation is exact. Its general solutions is
Example 2.1.44 Solve (e^{y} + 1)cos x dx + e^{y} sin x dy = 0.
Solution
(The equation is exact.)
Integrating, we get the general solution as
EXERCISE 2.4
Solve:
 (2x + 3y + 4)dx + (3x − 6y + 2)dy = 0.
Ans: x^{2} + 3xy − 3y^{2} + 4x + 2y + c = 0
(3x^{2} − 9x^{2}y^{2} + 2xy)dx + (6y^{2} − 6x^{3}y^{2} + x^{2})dy = 0.
Ans: x^{3} + 2y^{3} − 3x^{3}y^{2} + x^{2}y = c
e^{x}(sin x + cos x)sec y dx + e^{x}sin x sec y tan y dy = 0.
Ans: e^{x} sin x sec y = c
(cos x cos y − cot x)dx = sin x sin y dy.
Ans: sin x cos y = log c sin x

Ans: xy + y log x + x sin y = c
4y sin2xdx + (2y + 3 − 4cos^{2} x)dy = 0.
Ans: y^{2} + y − 2y cos2x = c

Ans:
 (x^{2} − ay)dx = (ax − y^{2})dy.
Ans: x^{3} + y^{3} = 3axy + c
xy^{2} = ce^{y2}
Ans:

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

Ans:

Ans:
2.1.5 Inexact Equation—Reducible to Exact Equation by Integrating Factors
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 ydx − xdy = 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.

 ydx + xdy + xy(ydx − xdy) = 0
EXERCISE 2.5
Find the integrating factors by inspection and solve the following:
 xdy − (y − 3x^{2})dx = 0.
Ans:
 ye^{−x} + e^{−x}dy = 2xy^{2}dx.
Ans:
 xdy + 2ydx = 2y^{2}xdy.
Ans:
 (x^{2} + y^{2})(xdy − ydx) = (x^{2} − y^{2})(xdx + ydy).
Ans: (x + y) = c(x − y)(x^{2} + y^{2})
2xydy − y^{2}dx + x^{3}e^{x}dx = 0.
Ans: y2 + x(x − 1)e^{x} + 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 ydx − xdy = 0 whose general solution is
Rule 2 If Mdx + Ndy = 0 is of the form
and Mx − Ny ≠ 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 = Py − Q 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
where a,b,c,d,m,n,p,q are all constants such that mq − np ≠ 0 then x^{h} y^{k} 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 (x^{2} + y^{2})dx − xydy = 0 by finding an integrating factor.
Solution This is a homogeneous differential equation.
Here,
Since , the equation is not exact
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 + x^{2} + y^{2}) dx − x^{2}dy = 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
By Rule 2, the integrating factor is if Mx − Ny ≠ 0.
Multiplying by the integrating factor the equation can be written as
Example 2.1.51 Solve (x^{2}y^{2} + xy + 1) ydx + (x^{2}y^{2} − xy + 1) xdy = 0.
Solution The equation is of the form
Integrating factor
Multiplying the equation by and omitting the constant,
The general solution is .
Example 2.1.52 Solve (x − y)dx − dy = 0.
Solution
(Constant can be considered as a function of x.)
By Rule 3, the integrating factor is .
Multiplying by the integrating factor = e^{x} the equation can be written as
Example 2.1.53 Solve (3xy − 2y^{2})dx + (x^{2} − 2xy)dy = 0.
Solution
By Rule 3, the integrating factor is .
Multiplying by integrating factor = x, the equation becomes
is the general solution of the differential equation.
Example 2.1.54 Solve (xy^{3} + y)dx + 2(x^{2}y^{2} + 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 + xy^{2})dx − xdy = 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 xy^{3} (ydx + 2xdy) + (3ydx + 5xdy) = 0.
Solution The given differential equation is
Here M = xy^{4} + 3y N = 2x^{2}y^{3} + 5x
M_{y} = 4xy^{3} + 3 N_{x} = 4xy^{3} + 5
Since M_{y} ≠ N_{x}, 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 x^{h}y^{k}, 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 x^{2} y^{4}. 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) + 8y^{4} (ydx + xdy) = 0.
Solution The given differential equation is
We observe that an integrating for this equation is of the form x^{h} y^{k} for some h, k Multiplying Eq. (2.66) by x^{h} y^{k}, it can be written in the form
where M = 3x^{h+1}y^{k+1} + 8x^{h}y^{k+5},
N = 2x^{h+2}y^{k} + 8x^{h+1}y^{k+4}
Exactness condition is
Now the lefthand side expression of Eq. (2.66) is exact
which is the required solution.
Example 2.1.58 Solve 2x^{2}(ydx + xdy) + y(ydx − xdy) = 0.
Solution The given differential equation is
Here  M = 2x^{2}y + y^{2}  N = 2x^{3} − xy 
∴  M_{y} = 2x^{2} + 2y  N_{x} = 6x^{2} − 2 y 
Since M_{y} ≠ N_{x}, the equation is not exact.
To find an integrating factor of the form x^{h}y^{k}, we multiply Eq. (2.70) by x^{h}y^{k} so that new value of
M = 2x^{h+2}y^{k+1} + x^{hyk+2} and
N = 2x^{h+3}y^{k} − x^{h+1}y^{k+1}
M_{y} = 2(k + 1)x^{h+2}y^{k} + (k + 2)x^{h}y^{k+1}
N_{x} = 2(h + 3)x^{h+2}y^{k} − (h + 1)x^{h}y^{k+1}
Equating M_{y} = N_{x}, we have
Integrating, we get
which is the required solution.
Example 2.1.59 Solve xy(ydx + xdy) + x^{2}y^{2}(2ydx − xdy) = 0.
Solution Multiplying the equation by an integrating factor x^{h} y^{k}, 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:
 (x^{2}y − 2xy^{2})dx − (x^{3}−3x^{2}y)dy = 0.
(Madras 1975, Karnataka 1971, Calcutta Hon. 1975)
Ans: (x/y) + log(y^{3}/x^{2}) = c
(x^{2} + y^{2})dx − 2xydy = 0.
Ans: x^{2} − y^{2} = cx
x^{2}ydx − (x^{3} + y^{3})dy = 0.
Ans:
 y(xy + 2x^{2}y^{2})dx + x(xy − x^{2}y^{2})dy = 0.
(Rajasthan 1969, Kanpur 1974, Karnataka 1971, Punjab 1971)
Ans:
 (x^{3}y^{3} + x^{2}y^{2} + xy + 1)ydx + (x^{3}y^{3} − x^{2}y^{2} − xy + 1)xdy = 0.
(Kanpur 1980, Gorakhpur 1972)
Ans:
 (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
(x^{2} + y^{2} + 2x)dx + 2ydy = 0.
(Srivenkateswara 1984, Calicut 1983)
Ans: e^{x}(x^{3} + y^{3}) = c
(y + log x)dx − xdy = 0.
(Marathawada 1994)
Ans: ex + y + log x + 1 = 0
 (3x^{2}y^{4} + 2xy)dx + (2x^{3}y^{3} − x^{2})dy = 0.
(Calcutta Hon. 1952, 1954; Utkal 1980)
Ans:
 (xy^{3} + y)dx + 2(x^{2}y^{2} + x + y^{4})dy = 0.
Ans: 3x^{2}y^{4} + 6xy^{2} + 2y^{6} = c
x(4ydx + 2xdy) + y^{3}(3ydx + 5xdy) = 0.
Ans: x^{4}y^{2} + x^{3}y^{3} = c
 xy^{3}(ydx + 2xdy) + (3ydx + 5xdy) = 0.
Ans:
2.1.6 Linear Equations
A differential equation of the form
where p_{0}, p_{1} and p_{2} are continuous functions of x in some interval I, is called a first order linear differential equation in y. Dividing Eq. (2.73) by p_{0} 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 nonhomogeneous 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 = Py − Q,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
Write the equation in the standard form
Find the integrating factor and multiply the equation by the integrating factor.
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 nonhomogeneous 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(P_{0}y = Qdx which, on integration, yields the general solution
Example 2.1.62
Solution This can be written as
Separating the variables and integrating
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 y^{2}dx + (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 (e^{x} − cos x)
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) = sec^{2} x, Q(x) = tan x sec^{2} x. This is linear in y. Also,
Integrating factor = e^{tanx}
The general solution is
Put tan x = t, sec^{2} xdx = dt
Integrating by parts
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
which, on integration, yields the general solution
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
Example 2.1.72 Solve
Solution This is a linear equation in y with
But, as it is, it can be written as
Integrating, the general solution is
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 x^{2}
Example 2.1.74 Solve
Solution Divide by x^{2}
Example 2.1.74 Solve
Solution Divide by x^{2}
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:

Ans:

Ans: xy sec x = tan x + C

Ans: y = Ce^{−2x} + e^{x} sin 2x
y′ + ytan x = sin 2x, y(0) = 1
[Hint: y sec x = sin^{2} x + c]
or y = cos x − cos^{3} x + ccos x
y(0) = 1 ⇒ 1 = c.
Ans: y = 2cos x − cos^{3} x
dx = (x + y + 1)dy.
Ans: x + y + 2 = ce^{y}
y′ + ycot x = 2x cosec x.
Ans: y = (x^{2} + c)cosec x

Ans: 16x^{2}y + x^{4}(1 − 4log x) = c

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

Ans:

Ans:

Ans: x sec y = tan y + c
ye^{y}dx = (y^{3} + 2xe^{y}dy).
Ans:

Ans:

Ans:

Ans:

[Hint: ]
on multiplication by sec^{2} x
⇒ ∫ d (y tan x) = ∫ xe^{x}dx.
Ans: y tan x = c + (x − 1)e^{x}

Ans: y = tan x − 1 + ce^{−tan x}

Ans: yx sec x = tan x + c

Ans: y log x = c + (log x)^{2}

Ans:

Ans: sin y = (1 + x)(e^{x} + c)

Ans: 2x = e^{y}(1 + x^{2})

Ans:
 (1 + y + x^{2}y)dx + (x + x^{2})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 = y^{1−a}.
Multiplying Eq. (2.91) by (1 − a)y^{−a}, 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.
Multiplying the equation by , we have
Integrating factor
The general solution is
Example 2.1.81 Solve y(2xy + e^{x})dx = e^{x}dy.
Solution
This is Bernoulli's equation.
Multiplying the equation by , we have
This is a linear equation in
Integrating factor = e^{∫1dx} = e^{x}
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:

Ans: (5 + cx^{2})x^{3}y^{5} = 2

Ans:

Ans: (log y)^{−1} = 1 + cx

Ans: y^{2} − x^{2} = cx − 1
 x(x − y)dy + y^{2}dx = 0.
Ans:

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

Ans: 2 (tan x + sec x) = y(2sin x + 1)
 y (2xy + ex)dx = e^{x}dy.
Ans: e^{x} = y(c − x^{2})

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

Ans: sin y = (1 + x)(e^{x} + c)

Ans:

Ans:

Ans:
 y(2xy + e^{x})dx − e^{x}dy = 0.
Ans: e^{x} + yx^{2} + cy = 0

Ans:

Ans:
2.2 APPLICATIONS OF ORDINARY DIFFERENTIAL EQUATIONS
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 (T_{0}).
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 T_{i} 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
 Identify T_{0}, the temperature of the surrounding medium. Then the general solution is given by Eq. (2.99).
 Use two given conditions and find the constant of integration c and the proportionality constant k.
 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 T_{0} = 30
Initial temperature T = T_{t} = 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 T_{0} = 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
 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
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
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
Uranium disintegrates at a rate proportional to the amount present at any instant. If m_{1} and m_{2} gms of uranium are present at time t_{1} and t_{2}, respectively. Show that the halflife of uranium is [(t_{2} − t_{1})log 2]log(m_{1} / m_{2}).
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:
The rate of decay of radium varies as its mass at a given time. Given that halflife of radium is 1600 yrs, find out the percentage of the mass of radium it will disintegrate in 200 yrs.
Ans: 8.3% approximately
2.2.1 Geometrical Applications
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 selforthogonal.
 In the electrical field, the paths along which the current flows are the orthogonal trajectories of the equipotential curves.
 In fluid mechanics, the stream lines and the equipotential lines are orthogonal trajectories of one another.
 In thermodynamics, the lines of heat flow are perpendicular to isothermal curves.
Method for finding orthogonal trajectories
(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 m_{1} be the slope of the curve of the given family and m_{2} 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 x^{2} – y^{2} = 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 yaxis.
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
where ‘c’ is a constant.
Example 2.2.8 Find the orthogonal trajectories of the family of semicubical parabolas ay^{2} = x^{3} 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
where ‘c’ is an arbitrary constant.
Example 2.2.9 Show that the system of confocal ellipses is selforthogonal.
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 y^{2} = 4a(x + a) is selforthogonal.
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 y_{1} 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 subtangent), 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 r^{n}sin nθ = a^{n} where a^{n} 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