4. Differential Equations of the First Order but not of the First Degree – Differential Equations

4

Differential Equations of the First Order but not of the First Degree

We will now discuss the solutions of differential equations which are of the first order but are of degree higher than one. Such differential equations will contain only the first order differential coefficient but will occur in a degree higher than one. It is usual to denote by p. The general form of such a differential equation is then

 

pn + A1 pn−1 + A2 pn−2 + … + An−1 p + An = 0

 

where A1, A2An are some functions of x and y.

Now we will consider various methods of solving the differential equations of the above type.

4.1 EQUATIONS SOLVABLE FOR p

Suppose the differential equation of first order and of degree n > 1 can be solved for p. That is, it can be resolved into n linear factors in p of the type

 

[pf, (x, y)] [pf2 (x, y)] … [pfn (x, y)] = 0

 

We can equate each factor to zero and the resulting differential equations of the first order and first degree can be solved. Let their solutions be

 

φ1 (x, y, c1) = 0, φ2 (x, y, c2) = 0, … φn (x, y, cn) = 0

 

where c1, c2,…cn are arbitrary constants. Without loss of generality we can replace the arbitrary constants c1, c2,…cn by a single arbitrary constant c because in any of the above n solutions c is free to take any real value. Thus the n solutions of the given differential equation are

 

φ1 (x, y, c) = 0, φ2 (x, y, c) = 0, …., φn (x, y, c) = 0

 

Combining the above equations we get a single composite solution as

 

φ1 (x, y, c) = φ2 (x, y, c) = …. φn (x, y, c) = 0

Example 4.1.1   Solve p2 − 5p + 6 = 0.

Solution   Resolving into linear factors the given differential equation can be written as

 

(p − 3) (p − 2) = 0

 

Its component equations are p = 3, p = 2.

Solving p = 3 i.e. we get y = 3x + c. Also, the solution of p = 2 i.e. is y = 2x + c.

So, the solutions of the given differential equation are

 

y = 3x + c, y = 2x + c

 

The single combined solution is

 

(y − 3xc) (y − 2xc) = 0

Example 4.1.2   Solve p (py) = x (x + y).

Solution   The given differential equation is

 

The first equation gives

The second equation is a linear equation

 

Integrating factor = ex

 

Its solution is yex = c + ∫xex dx = cxexex

 

or                                         y = cexx − 1

 

The single combined solution is

 

(2y + x2c) (y + x + 1 − cex) = 0

Example 4.1.3   Solve 2p2 + p − 1 = 0.

Solution   2p2 + p − 1 = (2p − 1) (p + 1)

 

 

Integrating we get

 

2y = x + c,     y = −x + c

 

combined solution is

 

(2yxc) (y + xc) = 0

Example 4.1.4   Solve p2 − 5px + 6x2 = 0.

Solution   p2 − 5px + 6x2 = (p − 3x) (p − 2x) = 0

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.5   Solve p2 − 5px − 6x2 = 0.

Solution   p2 − 5px − 6x2 = (p − 6x) (p + x) = 0

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.6   Solve (2p − 3)2 − (p − 2)2 = 0.

Solution   (2p − 3)2 − (p − 2)2 = (3p − 5) (p − 1) = 0

 

 

Integrating we get

 

3y = 5x + c,     y = x + c

 

combined solution is

 

(3y − 5xc) (yxc) = 0

Example 4.1.7   Solve (3px)2 − (p − 2x)2 = 0.

Solution   (3px)2 − (p − 2x)2 = (4p − 3x) (2p + x) = 0

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.8   Solve p2 + 2xp − 3x2 = 0.

Solution   p2 + 2xp − 3x2 = (p + x)2 − 4x2 = 0

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.9   Solve p2 − 2py cosh x + 1 = 0.

Solution   (p − cosh x)2 = −1 + cosh2x = sinh2x

(p − cosh x + sinh x) (p − cosh x − sinh x) = 0

 

 

Integrating we get

 

y = −ex + c,      y = ex + c

 

combined solution is

 

(y + exc) (yexc) = 0

Example 4.1.10   Solve yp2 + (xy) px = 0.

Solution   yp2 + (xy) px = yp (p −1) + x (p − 1) = 0

 

(p −1) = 0,     yp + x = 0

 

 

Integrating we get

 

y = x + c,     y2 + x2c2 = 0

 

combined solution is

 

(yxc) (y2 + x2c2) = 0

Example 4.1.11   Solve x + yp2 = p (1 + xy).

Solution   x + yp2 = p (1 + xy)

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.12   Solve xp2 + (yx) py = 0.

Solution   xp2 + (yx) py = 0

 

 

Integrating we get

 

y = x + c,     xy + c = 0

 

combined solution is

 

(yxc) (xy + c) = 0

Example 4.1.13   Solve p3ax4 = 0.

Solution   

Integrating we get

 

Example 4.1.14   Solve p2 + px + py + xy = 0.

Solution   p2 + px + py + xy = 0

 

 

Integrating we get

 

 

combined solution is

 

Example 4.1.15   Solve (p + y + x) (xp + y + x) (p + 2x) = 0.

Solution   (p + y + x) (xp + y + x) (p + 2x) = 0

 

linear equation; integrating factor is ex

Solution is    yex = c + ∫ − x ex dx

i.e.,                           yexcxex + ex

i.e.,                           y = cexx + 1

combined solution is

 

Example 4.1.16   Solve

Solution   

 

Separating the variables and integrating

 

 

or                           (x + c)2 + (yb)2 = 1

Example 4.1.17   Solve p (py) = x (x + y).

Solution   p (py) = x (x + y)

 

This is a linear equation

Integrating factor is ex

Solution is yex = c + ∫ xex dx = cxexex

Solution of

combined solution is

 

Example 4.1.18   Solve p2 + 2py cot x = y2.

Solution   p2 + 2py cot x = y2

 

 

The component equations are

 

 

 

Separating the variables and integrating (4.1) gives

 

 

Similarly (4.2) gives

 

 

Single combined solution is

 

Example 4.1.19   Solve 4y2p2 + 2xy (3x + 1) p + 3x3 = 0.

Solution   4y2p2 + 2xy (3x + 1) p + 3x3 = 0

 

 

Integrating

 

y2 + x3 + c = 0                       2y2 + x2 + c = 0

 

combined solution is

 

(y2 + x3 + c) (2y2 + x2 + c) = 0

Example 4.1.20   Solve xyp2 + (x2 + xy + y2) p + x2 + xy = 0.

Solution   xyp2 + x2p + xyp + x2 + y 2p + xy = 0

 

 

Integrating

 

y2 + x2 + c = 0                      2xy + x2 + c = 0

 

combined solution is

 

(y2 + x2 + c) (2xy + x2 + c) = 0
EXERCISE 4.1

Solve the following differential equations:

  1. p2 − 7p + 12 = 0.

    Ans: (y − 4xc) (y − 3xc) = 0

  2. y2 + xypx2p2 = 0.

    Ans:

  3. p2ax3 = 0.

    Ans: 25 (y + c)2 − 4ax5 = 0

  4. xyp2 − (x2 + y2) p + xy = 0.

    Ans: (x2y2c) (ycx) = 0

  5. x2p2 + xyp − 6y2 = 0.

    Ans: (yx3c) (ycx2) = 0

  6. xyp2 + p(3x2 − 2y2) − 6xy = 0.

    Ans: (ycx2) (y2 + 3x2c) = 0

  7. x2p2 − 2xyp + y2 = x2y2 + x4.

    Ans: y2x2 sinh2 (x + c) = 0

  8. yp2 + (xy)px = 0.

    Ans: (yxc) (x2 + y2c2) = 0

  9. x2p2 - 2xyp + 2y2 - x2 = 0.

    Ans:

  10. x + y p2 = (1 + xy) p.

    Ans:

4.2 EQUATIONS SOLVABLE FOR y

Suppose the given differential equation is solvable for y. Then it can be put in the form

 

 

Differentiating (4.3) w.r.t. x and denoting by p, we obtain

 

 

which is a differential equation in two variables x and p. Suppose it is possible to solve the differential equation (4.4). Let its solution be

 

 

where c is an arbitrary constant.

Eliminating p between (4.3) and (4.5) we get the required solution of (4.3) in the form ψ(x, y, c) = 0.

If it is not easily practicable to eliminate p between (4.3) and (4.5) we may solve (4.3) and (4.5) to get x and y in terms of p and c in the form x = f1 (p, c), y = f2 (p, c), which give us the required solution of (4.3) in the form of parametric equations the parameter being p.

Example 4.2.1   Solve y = 3x + a log p.

Solution   y = 3x + a log p

Differentiating w.r.t. x

 

 

Example 4.2.2   Solve p2py + 1 = 0.

Solution   

Differentiating w.r.t. x

 

 

Separating the variables

 

 

Integrating

 

 

The eliminant of p between the given equation and this equation is the solution.

Example 4.2.3   Solve y = x + a tan−1 p.

Solution   y = x + a tan−1 p

Differentiating w.r.t. x

 

 

Separating the variables

 

 

Putting into partial fractions and integrating

 

Example 4.2.4   Solve 3p5py + 1 = 0.

Solution   

Differentiating w.r.t. x we get

 

 

Separating the variables and integrating

 

 

 

The eliminant of p between the given equation and this equation is the solution.

Example 4.2.5   Solve y = p2x + p.

Solution   y = p2x + p

Differentiating w.r.t. x we get

 

 

Rewriting the equation as

 

 

This is a linear equation

Integrating factor

The solution is

 

 

multiplying the given equation by (p − 1)2 and using the value of x we get

 

(p − 1)2 y = p2 (cp + log p) + p ( p −1)2

Example 4.2.6   Solve y = p sin x + cos x.

Solution   y = p sin x + cos x

Differentiating w.r.t. x

 

 

This is a linear equation.

Integrating factor is

Solution is

 

 

This with the given relation is the solution.

Example 4.2.7   Solve y + px = p2 x4.

Solution   y + px = p2 x4

Differentiating w.r.t. x we get

 

 

Separating the variables of the second equation

 

 

Integrating we get

 

log p + 2 log x = log c
px2 = c

 

Eliminating p between this and 1− 2x3p = 0 we get the general solution as

Example 4.2.8   Solve y = 2xp + x2p4.

Solution   y = 2xp + x2p4

Differentiating w.r.t. x we get

 

 

 

Separating the variables and integrating

 

 

Eliminating p from this and the given relation. We get

 

 

The general solution is

 

(yc2)2 = 4cx

Example 4.2.9   Solve y + px = p2 x4.

Solution   y + px = p2 x4

 

y = − px + p2x4

 

Differentiating w.r.t. x and denoting by p we get

 

 

 

Separating the variables and integrating

 

 

we get              px2 = c

 

Substituting in the given relation

 

 

which is the required solution.

Example 4.2.10   Solve y = 2pxp2.

Solution   y = 2pxp2

Differentiating w.r.t. x

 

 

This is a linear equation

Integrating factor is

The solution is

 

Here it is not easily practicable to eliminate p between this and the given relation. So putting the value of x in the given relation

 

 

 

Equations (4.6) and (4.7) constitute the solution.

Example 4.2.11   Solve

Solution   yx = xp + p2 or y = x (1 + p) + p2

Differentiating w.r.t. x

 

 

Linear equation with integrating factor ep.

Its solution is

 

 

Substituting this value of x in the given relation

 

 

Equations (4.8) and (4.9) constitute the solution.

Example 4.2.12   Solve

Solution   

Differentiating w.r.t. x

 

 

 

Integrating we get

 

 

From the given relation y2 = a2 + a2p2

 

 

 

 

where c1 = ca log a

Hence the required solution is

 

EXERCISE 4.2

Solve the following differential equations:

  1. y = 2px + p4x2.

    Ans: (yc2)2 = 4cx

  2. 4p3 + 3xp = y.

    Ans:

  3. x2 + p2x = yp.

    Ans:

  4. y − 2px = xp2.

    Ans:

  5. p2xp + y = 0.

    Ans: y = cxc2

  6. x3p2 + x2yp + 4 = 0.

    Ans: cxy + 4x + c2 = 0

  7. 2xp3 − 6yp2 + x4 = 0.

    Ans: 2c3x3 = 1 − 6 c2y

  8. xyp = ap2.

    Ans:

4.3 EQUATIONS SOLVABLE FOR x

Suppose the given differential equation is solvable for x. Then it can be put in the form

 

 

Differentiating (4.11) w.r.t. y and writing for we get

 

 

which is a differential equation in two variables y and p. Suppose it is possible to solve the differential equation (4.12). Let its solution be

 

 

where c is the arbitrary constant.

Eliminating p between (4.11) and (4.12) we get the required solution of (4.11) in the form

 

 

which give us the required solution of (4.11) in the form of parametric equations, the parameter being p.

Example 4.3.1   Solve y = 2px + y2p3.

Solution   The given differential equation can be written as

 

 

Differentiating w.r.t. y we get

 

 

From p + y = 0 we have or py = c

Substituting the value of p i.e. in the given relation

 

 

or y2 = 2cx + c2 which is the required solution.

Example 4.3.2   Solve p3 − 4xyp + 8y2 = 0.

Solution   4xyp = p3 + 8y2 or

Differentiating w.r.t. y we get

 

 

 

The first equation gives p2 = cy.

From the differential equation, we have

 

8y2 = p (4xyp2)

 

Substituting in this equation

 

 

Hence the required solution is y = c (xc)2.

Example 4.3.3   Solve ap2 + pyx = 0.

Solution   x = py + ap2

Differentiating w.r.t. y and writing for

 

 

Linear differential equation. I.F. is

The solution is

 

 

 

 

Substituting this in the given differential equation we get

 

 

Equations (4.15) and (4.16) constitute the general solution.

Example 4.3.4   Solve xp3 = a + bp.

Solution   

Differentiating w.r.t. y

 

 

Integrating

 

 

The general solution is

 

Example 4.3.5   Solve y2 log y = xyp + p2.

Solution   

Differentiating w.r.t. y we get

 

 

Eliminating p between this and the given relation

 

y2 log y = cxy2 + c2y2 or log y = cx + c2

 

which is the general solution.

Example 4.3.6   Solve

Solution   Solving for x,

 

 

Differentiating w.r.t. y

 

 

Integrating

 

 

Equations (4.17) and (4.18) constitute the general solution.

Example 4.3.7   Solve ayp2 + (2xb) py = 0.

Solution   Solving for x,

Differentiating w.r.t. y

 

 

From the first factor

Integrating                                   py = c

Eliminating p between this and the original equation

 

 

i.e.                                 ac2 + (2xb) cy2 = 0

 

This is the solution.

Example 4.3.8   Solve 2px = 2 tan y + p3 cos2 y.

Solution   Solving for x

 

 

Differentiating w.r.t. y

 

separating the variables and integrating

log p = log sec y + log c or p = c sec y

 

Eliminating p between this and the original equation

 

 

which is the general solution.

EXERCISE 4.3

Solve the following differential equations:

  1. Ans:

  2. p2y + 2px = y.

    Ans: y2 = 2cx + c2

  3. y = 2px + y2p3.

    Ans: y2 = 2cx + c2

  4. 4xp2 + 4ypy4 = 0.

    Ans: 4c (cxy + 1) = y

  5. p3 − 4xyp + 8y2 = 0.

    Ans: c (c − 4x)2 = 64y

  6. p3 − (y + 3) p + x = 0.

    Ans:

  7. y2 log y = xyp + p2.

    Ans: log y = cx + c2

  8. 4 (xp2 + yp) = y4.

    Ans: y = 4c (xyc + 1)