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
where A_{1}, A_{2} … A_{n} 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
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
where c_{1}, c_{2},…c_{n} are arbitrary constants. Without loss of generality we can replace the arbitrary constants c_{1}, c_{2},…c_{n} 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
Combining the above equations we get a single composite solution as
Example 4.1.1 Solve p^{2} − 5p + 6 = 0.
Solution Resolving into linear factors the given differential equation can be written as
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
The single combined solution is
Example 4.1.2 Solve p (p − y) = x (x + y).
Solution The given differential equation is
The second equation is a linear equation
Its solution is ye^{−x} = c + ∫xe^{−x} dx = c − xe^{−x} − e^{−x}
or y = ce^{x} − x − 1
The single combined solution is
Example 4.1.3 Solve 2p^{2} + p − 1 = 0.
Solution 2p^{2} + p − 1 = (2p − 1) (p + 1)
Integrating we get
combined solution is
Example 4.1.4 Solve p^{2} − 5px + 6x^{2} = 0.
Solution p^{2} − 5px + 6x^{2} = (p − 3x) (p − 2x) = 0
Integrating we get
combined solution is
Example 4.1.5 Solve p^{2} − 5px − 6x^{2} = 0.
Solution p^{2} − 5px − 6x^{2} = (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
combined solution is
Example 4.1.7 Solve (3p − x)^{2} − (p − 2x)^{2} = 0.
Solution (3p − x)^{2} − (p − 2x)^{2} = (4p − 3x) (2p + x) = 0
Integrating we get
combined solution is
Example 4.1.8 Solve p^{2} + 2xp − 3x^{2} = 0.
Solution p^{2} + 2xp − 3x^{2} = (p + x)^{2} − 4x^{2} = 0
Integrating we get
combined solution is
Example 4.1.9 Solve p^{2} − 2py cosh x + 1 = 0.
Solution (p − cosh x)^{2} = −1 + cosh^{2}x = sinh^{2}x
(p − cosh x + sinh x) (p − cosh x − sinh x) = 0
Integrating we get
combined solution is
Example 4.1.10 Solve yp^{2} + (x − y) p − x = 0.
Solution yp^{2} + (x − y) p − x = yp (p −1) + x (p − 1) = 0
Integrating we get
combined solution is
Example 4.1.11 Solve x + yp^{2} = p (1 + xy).
Solution x + yp^{2} = p (1 + xy)
Integrating we get
combined solution is
Example 4.1.12 Solve xp^{2} + (y − x) p − y = 0.
Solution xp^{2} + (y − x) p − y = 0
Integrating we get
combined solution is
Example 4.1.13 Solve p^{3} − ax^{4} = 0.
Solution
Integrating we get
Example 4.1.14 Solve p^{2} + px + py + xy = 0.
Solution p^{2} + 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 e^{x}
Solution is ye^{x} = c + ∫ − x e^{x} dx
i.e., ye^{x} − c − xe^{x} + e^{x}
i.e., y = ce^{−x} − x + 1
combined solution is
Example 4.1.16 Solve
Solution
Separating the variables and integrating
or (x + c)^{2} + (y − b)^{2} = 1
Example 4.1.17 Solve p (p − y) = x (x + y).
Solution p (p − y) = x (x + y)
This is a linear equation
Integrating factor is e^{−x}
Solution is ye^{−x} = c + ∫ xe^{−x} dx = c − xe^{−x} − e^{−x}
Solution of
combined solution is
Example 4.1.18 Solve p^{2} + 2py cot x = y^{2}.
Solution p^{2} + 2py cot x = y^{2}
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 4y^{2}p^{2} + 2xy (3x + 1) p + 3x^{3} = 0.
Solution 4y^{2}p^{2} + 2xy (3x + 1) p + 3x^{3} = 0
Integrating
combined solution is
Example 4.1.20 Solve xyp^{2} + (x^{2} + xy + y^{2}) p + x^{2} + xy = 0.
Solution xyp^{2} + x^{2}p + xyp + x^{2} + y ^{2}p + xy = 0
Integrating
combined solution is
EXERCISE 4.1
Solve the following differential equations:
 p^{2} − 7p + 12 = 0.
 y^{2} + xyp − x^{2}p^{2} = 0.
Ans:
 p^{2} − ax^{3} = 0.
Ans: 25 (y + c)^{2} − 4ax^{5} = 0
 xyp^{2} − (x^{2} + y^{2}) p + xy = 0.
Ans: (x^{2} − y^{2} − c) (y − cx) = 0
 x^{2}p^{2} + xyp − 6y^{2} = 0.
Ans: (yx^{3} − c) (y − cx^{2}) = 0
 xyp^{2} + p(3x^{2} − 2y^{2}) − 6xy = 0.
Ans: (y − cx^{2}) (y^{2} + 3x^{2} − c) = 0
 x^{2}p^{2} − 2xyp + y^{2} = x^{2}y^{2} + x^{4}.
Ans: y^{2} − x^{2} sinh^{2} (x + c) = 0
 yp^{2} + (x − y)p − x = 0.
Ans: (y − x − c) (x^{2} + y^{2} − c^{2}) = 0
 x^{2}p^{2}  2xyp + 2y^{2}  x^{2} = 0.
Ans:
 x + y p^{2} = (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 = f_{1} (p, c), y = f_{2} (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 p^{2} − py + 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 3p^{5} − py + 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 = p^{2}x + p.
Solution y = p^{2}x + 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
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 = p^{2} x^{4}.
Solution y + px = p^{2} x^{4}
Differentiating w.r.t. x we get
Separating the variables of the second equation
Integrating we get
Eliminating p between this and 1− 2x^{3}p = 0 we get the general solution as
Example 4.2.8 Solve y = 2xp + x^{2}p^{4}.
Solution y = 2xp + x^{2}p^{4}
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
Example 4.2.9 Solve y + px = p^{2} x^{4}.
Solution y + px = p^{2} x^{4}
Differentiating w.r.t. x and denoting by p we get
Separating the variables and integrating
we get px^{2} = c
Substituting in the given relation
which is the required solution.
Example 4.2.10 Solve y = 2px − p^{2}.
Solution y = 2px − p^{2}
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 y − x = xp + p^{2} or y = x (1 + p) + p^{2}
Differentiating w.r.t. x
Linear equation with integrating factor e^{p}.
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 y^{2} = a^{2} + a^{2}p^{2}
where c_{1} = c − a log a
Hence the required solution is
EXERCISE 4.2
Solve the following differential equations:
 y = 2px + p^{4}x^{2}.
Ans: (y − c^{2})^{2} = 4cx
 4p^{3} + 3xp = y.
Ans:
 x^{2} + p^{2}x = yp.
 y − 2px = xp^{2}.
Ans:
 p^{2} − xp + y = 0.
Ans: y = cx − c^{2}
 x^{3}p^{2} + x^{2}yp + 4 = 0.
Ans: cxy + 4x + c^{2} = 0
 2xp^{3} − 6yp^{2} + x^{4} = 0.
Ans: 2c^{3}x^{3} = 1 − 6 c^{2}y
 x − yp = ap^{2}.
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 + y^{2}p^{3}.
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 y^{2} = 2cx + c^{2} which is the required solution.
Example 4.3.2 Solve p^{3} − 4xyp + 8y^{2} = 0.
Solution 4xyp = p^{3} + 8y^{2} or
Differentiating w.r.t. y we get
The first equation gives p^{2} = cy.
From the differential equation, we have
Substituting in this equation
Hence the required solution is y = c (x − c)^{2}.
Example 4.3.3 Solve ap^{2} + py − x = 0.
Solution x = py + ap^{2}
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 xp^{3} = a + bp.
Solution
Differentiating w.r.t. y
Integrating
The general solution is
Example 4.3.5 Solve y^{2} log y = xyp + p^{2}.
Solution
Differentiating w.r.t. y we get
Eliminating p between this and the given relation
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 ayp^{2} + (2x − b) p − y = 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. ac^{2} + (2x − b) c − y^{2} = 0
This is the solution.
Example 4.3.8 Solve 2px = 2 tan y + p^{3} cos^{2} y.
Solution Solving for x
Differentiating w.r.t. y
separating the variables and integrating
Eliminating p between this and the original equation
which is the general solution.
EXERCISE 4.3
Solve the following differential equations:

Ans:
 p^{2}y + 2px = y.
Ans: y^{2} = 2cx + c^{2}
 y = 2px + y^{2}p^{3}.
Ans: y^{2} = 2cx + c^{2}
 4xp^{2} + 4yp − y^{4} = 0.
Ans: 4c (cxy + 1) = y
 p^{3} − 4xyp + 8y^{2} = 0.
Ans: c (c − 4x)^{2} = 64y
 p^{3} − (y + 3) p + x = 0.
Ans:
 y^{2} log y = xyp + p^{2}.
Ans: log y = cx + c^{2}
 4 (xp^{2} + yp) = y^{4}.
Ans: y = 4c (xyc + 1)