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Differential Equations Introduction Solved Problems

Existence And Uniqueness Theorem Solved Problems

Example. 1: Solve : \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)

Solution.

Given equation is \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0\)…..(1)

Separating the variables: \(\frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}=0\)

⇒ \(\int \frac{d y}{\sqrt{1-y^2}}+\int \frac{d x}{\sqrt{1-x^2}}=c\) (c being arbitrary constant)

∴ The general solution of (1) is: \(\text{Sin}^{-1} y+\text{Sin}^{-1} x=c\)

Examples Of Solved Problems On Existence And Uniqueness Theorem 

Example. 2: Solve : \(y \frac{d y}{d x}=x e^{x^2+y^2}\)

Solution.

Given equation is \(y \frac{d y}{d x}=x e^{x^2} \cdot e^{y^2}\) …………………(1)

Separating the variables we have : \(x e^{x^2} d x=y e^{-y^2} d y \Rightarrow \int x e^{x^2} d x=\int y e^{-y^2} d y\)

Put \(x^2=u, y^2=t \Rightarrow 2 x d x=d u, 2 y d y=d t\)

∴ \(\frac{1}{2} \int e^u d u=\frac{1}{2} \int e^{-t} d t+\frac{c}{2} \Rightarrow e^u=-e^{-t}+c\)

∴ The general solution of (1) is \(e^{x^2}+e^{-y^2}=c\)

Application Of Existence And Uniqueness Theorem In Differential Equations

Example. 3: Solve : \(\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)\)

Solution.

Given \(\left(\frac{d y}{d x}\right) \tan y=2 \sin x \cos y\) ………………..(1)

Separating the variables : \(\frac{\tan y}{\cos y} d y=2 \sin x d x\)

=> \(\int\) sec y tan y dy = \(\int\) 2 sin x dx => \(\sec y=-2 \cos x+c\)

∴ The general solution of (1) is sec y + 2 cos x = c

Solved Examples Of Initial Value Problems Using The Existence And Uniqueness Theorem

Example. 4: Solve: \(\log \left(\frac{d y}{d x}\right)=a x+b y\)

Solution:

Given equation is \(\log \left(\frac{d y}{d x}\right)=a x+b y\) ………………………(1)

⇒ \(\frac{d y}{d x}=e^{a x+b y}=e^{a x} \cdot e^{b y} \Rightarrow e^{a x} d x=e^{-b y} d y\)

Separating the variables => \(\int e^{a x} d x=\int e^{-b y} d y+\mathrm{c} \Rightarrow \frac{e^{a x}}{a}=\frac{e^{-b y}}{-b}+\mathrm{c}\)

∴ The given solution of (1) is \(b e^{a x}+a e^{-b y}=c\)

Problems On Existence And Uniqueness Theorem With Solutions 

Example. 5: Solve : \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\)

Solution:

Given equation is \(\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}\) ……………………..(1)

Separating the variables : x(2 log x +1) dx = (sin y + y cos y) dy

Integrating: \(2 \int x \log x d x+\int x d x=\int \sin y d y+\int y \cos y d y\)

⇒ \(2\left(\frac{x^2}{2} \log x-\int \frac{x^2}{2} \cdot \frac{1}{x} d x\right)+\frac{x^2}{2}=-\cos y+y \sin y-\int \sin y d y\)

⇒ \(x^2 \log x-\frac{x^2}{2}+\frac{x^2}{2}=-\cos y+y \sin y+\cos y+c \Rightarrow x^2 \log x=y \sin y+c\)

The general solution of (1) is \(x^2 \log x=y \sin y+c\)

Lipschitz Condition In Existence And Uniqueness Theorem Problems

Example. 6: Solve \(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\)

Solution:

Given equation is \(y-x \frac{d y}{d x}=a\left(y^2+\frac{d y}{d x}\right)\) ………………….(1)

⇒ \((a+x) \frac{d y}{d x}=y-a y^2\)

Separating the variables : \(\frac{1}{y-a y^2} d y=\frac{1}{a+x} d x \Rightarrow \int \frac{1}{y(1-a y)} d y=\int \frac{1}{a+x} d x+c_1\)

⇒ \(\int\left(\frac{1}{y}+\frac{a}{1-a y}\right) d y=\int \frac{1}{a+x} d x+c_1 \Rightarrow \log |y|-\log |1-a y|=\log \mid a+x \mid+\log c\)

⇒ \(\log \left|\frac{y}{1-a y}\right|=\log |c(a+x)| \Rightarrow \frac{y}{1-a y}=c(a+x)\)

∴ The general solution of (1) is y = c (a + x) (1 – ay)

Existence And Uniqueness Theorem Numerical Examples

Example. 7: Solve : \(3 e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0\)

Solution.

Given equation is \(3 e^x \tan y d x+\left(1-e^x\right) \sec ^2 y d y=0\) …………………..(1)

Separating the variables: \(\frac{3 e^x}{1-e^x} d x+\frac{\sec ^2 y}{\tan y} d y=0\)

⇒ \(3 \int \frac{e^x}{1-e^x} d x+\int \frac{\sec ^2 y}{\tan y} d y=\log c \quad \Rightarrow-3 \log \left|\left(1-e^x\right)\right|+\log |\tan y|=\log c\)

⇒ \(\log \left|\left(1-e^x\right)^{-3}\right|+\log |\tan y|=\log c \Rightarrow \log \left|\frac{\tan y}{\left(1-e^x\right)^3}\right|=\log c \Rightarrow \frac{\tan y}{\left(1-e^x\right)^3}=c\)

∴ The general solution of (1) is \(\tan y=c\left(1-e^x\right)^3\)

Differential Equations Problems Explained Using Existence And Uniqueness Theorem

Example. 8: Solve : \(\frac{d y}{d x}=(4 x+y+1)^2\)

Solution:

Given equation is \(\frac{d y}{d x}=(4 x+y+1)^2\)……..(1)

Let \(4 x+y+1=u \Rightarrow 4+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-4\)…..(2)

From (1) and (2): \(\frac{d u}{d x}-4=u^2 \Rightarrow \frac{d u}{d x}=u^2+4\).

Separating the variables: \(\frac{d u}{u^2+4}=d x\)

⇒ \(\int \frac{d u}{u^2+2^2}=\int d x+\frac{c}{2} \Rightarrow \frac{1}{2} \text{Tan}^{-1}\left(\frac{u}{2}\right)\)

= \(x+\frac{c}{2} \Rightarrow \text{Tan}^{-1}\left(\frac{u}{2}\right)=2 x+c\)

u = \(2 \tan (2 x+c)\)

∴ The General solution is \(4 x+y+1=2 \tan (2 x+c)\)

(because \(u=4 \dot{x}+y+1)\)

Worked Examples On Existence And Uniqueness Theorem In Differential Equations

Example 9: Solve : \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\)

Solution:

Given equation is \(\frac{d y}{d x}=\cos (x+y)+\sin (x+y)\) …………………..(1)

Let \(x+y=u \Rightarrow 1+\frac{d y}{d x}=\frac{d u}{d x} \Rightarrow \frac{d y}{d x}=\frac{d u}{d x}-1\) ……………………..(2)

(1)and (2) => \(\frac{d u}{d x}-1=\cos u+\sin u \Rightarrow \frac{d u}{d x}=(1+\cos u)+\sin u\)

⇒ \(\frac{d u}{d x}=2 \cos ^2 \frac{u}{2}+2 \sin \frac{u}{2} \cos \frac{u}{2}=2 \cos ^2 \frac{u}{2}\left(1+\tan \frac{u}{2}\right)\)

Separating the variables : \(\frac{d u}{2 \cos ^2(u / 2)[1+\tan (u / 2)]}=d x\)

⇒ \(\int \frac{\sec ^2(u / 2)}{2[1+\tan (u / 2)]} d u=\int d x+c \Rightarrow \log \left(1+\tan \frac{u}{2}\right)=x+c\)

∴ The General Solution is \(\log \left[1+\tan \left(\frac{x+y}{2}\right)\right]=x+c\)

Example. 10: solve : \(\frac{d y}{d x}-x \tan (y-x)=1\)

Solution:

Given equation is \(\frac{d y}{d x}-x \tan (\cdot y-x)=1\) …………………..(1)

Put \(y-x=z \Rightarrow \frac{d y}{d x}-1=\frac{d z}{d x} \Rightarrow \frac{d y}{d x}=\frac{d z}{d x}+1\) ……………………(2)

(1)and(2) => \(\frac{d z}{d x}+1-x \tan z=1 \Rightarrow \frac{d z}{d x}=x \tan z\)

Separating the variables: \(\frac{d z}{\tan z}=x d x \Rightarrow \int \cot z d z=\int x d x+c \Rightarrow \log |\sin z|=\frac{x^2}{2}+\frac{c}{2}\)

∴ The general solution of (1) is \(2 \log |\sin (y-x)|=x^2+c\)

Example. 11: Solve : \(\left(\frac{x+y+a}{x+y-b}\right) \frac{d y}{d x}=\frac{x+y-a}{x+y+b}\)

Solution:

Given equation is

Put x + y = z in the given equation => \(1+\frac{d y}{d x}=\frac{d z}{d x}\)

∴ Given differential equation becomes \(\left(\frac{z+a}{z-b}\right)\left(\frac{d z}{d x}-1\right)=\frac{z-a}{z+b}\)

⇒ \(\frac{d z}{d x}=1+\frac{(z-a)(z-b)}{(z+a)(z+b)}=\frac{z^2+z(a+b)+a b+z^2-z(a+b)+a b}{(z+a)(z+b)}\)

⇒ \(\frac{d z}{d x}=\frac{2\left(z^2+a b\right)}{z^2+z(a+b)+a b}\)

Separating the variables: \(\Rightarrow \frac{z^2+z(a+b)+a b}{z^2+a b} d z=2 d x\)

⇒ \(\int\left[\frac{z^2+a b}{z^2+a b}+\frac{z(a+b)}{z^2+a b}\right] d z=\int 2 d x+c\)

⇒ \(\int d z+\frac{a+b}{2} \int \frac{2 z}{z^2+a b} d z=2 x+c \Rightarrow z+\frac{a+b}{2} \log \left|z^2+a b\right|=2 x+c\)

∴ The general solution is \(x+y+\frac{a+b}{2} \log \left|(x+y)^2+a b\right|=2 x+c\)

Example. 12: Find the equation of the curve passing through the point (1,1) whose differential equation is (y- yx) dx + (x+xy) dy = 0.

Solution :

Given equation is (y- yx) dx + (x+xy) dy = 0 => y (1 – x) dx + x (1+ y) dy =0

Separating the variables: \(\frac{1-x}{x} d x+\frac{1+y}{y} d y=0 \Rightarrow\left(\frac{1}{x}-1\right) d x+\left(\frac{1}{y}+1\right) d y=0\)

⇒ \(\int\left(\frac{1}{x}-1\right) d x+\int\left(\frac{1}{y}+1\right) d y=c \Rightarrow \log |x|-x+\log |y|+y=c\)

Given the curve passes through the point (1,1): = log 1 -1+ log 1 +1 = c => c = 0

∴ The equation of the curve passing through the point (1,1) is \(\log |x y|+y-x=0 \Rightarrow x y=e^{x-y}\)

Differential Equations of First Order But Not of First Degree Differential Equations Solvable For X

Let f(x,y,p) = 0 be the given differential equation ……………………(1)

If equation (1) cannot be split up into rational and linear factors and (1) is of first degree in x, then (1) can be solved for x.

(1) can be expressed in the form x = F(y,p) ……………………….(2)

Differentiating (2) w.r.t. y gives an equation of the form \(\frac{1}{p}=g\left(y, p, \frac{d p}{d y}\right)\) …………………(3)

Since (3) is an equation in two variables p and y , it can be solved.

∴ The solution of (3) is \(\phi(y, p, c)=0\) …………………(4).

Eliminating p from (1) and (4), general solution of (1) is \(\psi(x, y, c)=0\).

Note 1. If it is not possible to eliminate p, then the values of x and y in terms of p in the form \(x=f_1(p, c)\) and \(y=f_2(p, c)\) together give the general solution.

2. This method is especially useful for equations y being absent.

Differential Equations of First Order But Not of First Degree Solved Problems

Example. 1. Solve \(y^2 \log y=x p y+p^2\)
Solution.

Given equation is \(y^2 \log y=x p y+p^2\) …………………………(1)

Since x is the first degree in (1), it can be solved for x. \(x=\frac{y \log y}{p}-\frac{p}{y}\) ………………………(2)

Differentiating (2) w.r.t. y: \(\frac{1}{p}=(1+\log y) \frac{1}{p}-\frac{y \log y}{p^2} \frac{d p}{d y}-\frac{1}{y} \frac{d p}{d y}+\frac{p}{y^2}\)

⇒ \(\frac{1}{p}=\frac{1}{p}+\frac{1}{\mathrm{p}} \log y+\frac{p}{y^2}-\left(\frac{y \log y}{p^2}+\frac{1}{y}\right) \frac{d p}{d y}\)

⇒ \(\frac{p}{y}\left(\frac{y \log y}{p^2}+\frac{1}{y}\right)-\left(\frac{y \log y}{p^2}+\frac{1}{y}\right) \frac{d p}{d y}=0\)

⇒ \(\left(\frac{y \log y}{p^2}+\frac{1}{y}\right)\left(\frac{p}{\mathrm{y}}-\frac{d p}{d y}\right)=0 \Rightarrow \frac{y \log y}{p^2}+\frac{1}{y}=0\) ………………………(3)

∴ \(\frac{p}{y}-\frac{d p}{d y}=0\) ……………………….(4)

(3) is discarded as it gives a singular solution

Solving(4) : \(\frac{d p}{d y}=\frac{p}{y} \Rightarrow \frac{d p}{p}=\frac{d y}{y} \Rightarrow \int \frac{d p}{p}=\int \frac{d y}{y}+\log c\)

⇒ \(\log p=\log y+\log c \Rightarrow \log p=\log c y \Rightarrow p=c y\) …………………………(5)

Eliminating p from (1) and (5): \(y^2 \log y=c x y^2+c^2 y^2 \Rightarrow \log y=c x+c^2\)

∴ The general solution of (1) is \(\log y=c x+c^2\)

Differential Equations Of First Order But Not First Degree

Example. 2. Solve \(x=y+p^2\)
Solution.

Given equation is \(x=y+p^2\) ………………………….(1)

Differentiating (1) w.r.t. \(y \Rightarrow \frac{d x}{d y}=1+2 p \frac{d p}{d y} \Rightarrow \frac{d p}{d y}=\frac{1-p}{2 p^2}\)

Separating the variables : \(d y=\frac{-2 p^2}{p-1} d p \Rightarrow \int d y=\int \frac{-2 p^2}{p-1} d p+c\)

⇒ \(y=-2 \int\left(p+1+\frac{1}{p-1}\right) d p+c \Rightarrow-2\left[\frac{1}{2} p^2+p+\log (p-1)\right]+c\) ………………………..(2)

Substituting the value of y from (2) in (1), we get: \(x=c-2[p+\log (p-1)]\) ………………………(3)

which shows that it is not possible to eliminate p from (1) and (2).

∴ The general solution of(1) is \(x=c-2[p+\log (p-1)], y=-p^2-2 p-2 \log (p-1)+c\)

Differential Equations Solvable For x Explained

Example. 3. Solve \(a y p^2+(2 x-b) p-y=0\) where p = dy / dx, a and h are real numbers.

Solution.

Given equation is \(a y p^2+(2 x-b) p-y=0\)

Since x is of first degree in (1), it can be solved for x. \(\Rightarrow 2 x=\frac{y}{p}-a y p+b\)

Differentiating w.r.t.y: \(\frac{2}{p}=\frac{1}{p} \cdot 1-\frac{y}{p^2} \frac{d p}{d y}-a\left(p+y \frac{d p}{d y}\right)\)

⇒ \(\frac{1}{p}+\frac{y}{p^2} \frac{d p}{d y}+a\left(p+y \frac{d p}{d y}\right)=0\)

⇒ \(\frac{1}{p^2}\left(p+y \frac{d p}{d y}\right)+a\left(p+y \frac{d p}{d y}\right)=0 \Rightarrow\left(p+y \frac{d p}{d y}\right)\left(\frac{1}{p^2}+a\right)=0\)

∴ \(p+y \frac{d p}{d y}=0\) ……………………..(2)

∴ \(\frac{1}{p^2}+a=0\) …………………….(3)

(3) is discarded as it gives a singular solution.

Solving (2): \(p+y \frac{d p}{d y}=0 \Rightarrow \frac{d p}{p}+\frac{d y}{y}=0\)    (separating variables)

Integrating : \(\int \frac{d p}{p}+\int \frac{d y}{y}=\log c \Rightarrow \log p+\log y=\log c\)

⇒ \(\log p y=\log c \Rightarrow p y=c \Rightarrow p=c / y\)………………….(4)

By eliminating p from (1) and (4) => \(\text { ay }\left(\frac{c^2}{y^2}\right)+(2 x-b) \frac{c}{y}-y=0\)

∴ The general solution of (1) is \(a c^2+(2 x-b) c-y^2=0\)

Examples Of Differential Equations Solvable For x 

Example. 4. Solve \(a p^2+p y-x=0\)
Solution.

Given equation is \(x=y p+a p^2\) ……………………..(1)

Differentiating (1) w.r.t.y \(\Rightarrow \frac{d x}{d y}=p+y \frac{d p}{d y}+2 a p \frac{d p}{d y}\)

⇒ \(\frac{1}{p}-p=(y+2 a p) \frac{d p}{d y} \Rightarrow \frac{d y}{d p}=\frac{p y+2 a p^2}{1-p^2}\)

⇒ \(\frac{d y}{d p}-\frac{p}{1-p^2} y=\frac{2 a p^2}{1-p^2}\) which is linear in y and p ……………………….(2)

∴ \(\text { I.F. }=\exp \left(\int \frac{-p}{1-p^2} d p\right)=\exp \left[\frac{1}{2} \log \left(1-p^2\right)\right]=\sqrt{1-p^2}\)

The general solution of (2) is \(y \sqrt{1-p^2}=2 a \int \frac{p^2}{1-p^2}, \sqrt{1-p^2} d p+c\)

⇒ \(y \sqrt{1-p^2}=2 a \int \frac{1-\left(1-p^2\right)}{\sqrt{1-p^2}} d p+c=2 a \int \frac{d p}{\sqrt{1-p^2}}-2 a \int \sqrt{1-p^2} d p+c\)

⇒ \(y \sqrt{1-p^2}=2 a \sin ^{-1} p-2 a\left[\frac{p}{2} \sqrt{1-p^2}+\frac{1}{2} \sin ^{-1} p\right]+c\)

= \(a \sin ^{-1} p-a p \sqrt{1-p^2}+c \Rightarrow y=\frac{a \sin ^{-1} p+c}{\sqrt{1-p^2}}-a p\) ……………………….(3)

By eliminating y from (1) and (3) \(\Rightarrow x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\) …………………….(4)

∴ (3) and (4) together form the general solution of (1).

First-Order Equations Not Of First Degree With Solutions 

Example. 5. Solve \(\)
Solution.

Given \(p^3-4 x y p+8 y^2=0 \Rightarrow 4 x=\frac{p^2}{y}+\frac{8 y}{p}\) …………………………(1)

Differentiating (1) w.r.t.y \(\Rightarrow \frac{4}{p}=\frac{2 p}{y} \frac{d p}{d y}-\frac{p^2}{y^2}+\frac{8}{p}-\frac{8 y}{p^2} \frac{d p}{d y}\)

⇒ \(\frac{p^2}{y^2}-\frac{4}{p}=\left(\frac{2 p}{y}-\frac{8 y}{p^2}\right) \frac{d p}{d y} \Rightarrow \frac{p^3-4 y^2}{p y^2}=\frac{2\left(p^3-4 y^2\right)}{p^2 y} \frac{d p}{d y}\)

⇒ \(\left(p^3-4 y^2\right)\left(\frac{1}{y}-\frac{2}{p} \frac{d p}{d y}\right)=0 \Rightarrow p^3-4 y^2=0\) …………………….(2)

\(\frac{1}{y}-\frac{2}{p} \frac{d p}{d y}=0\) …………………….(3)

(2) is discarded as it gives a singular solution.

Solving (3): \(\frac{d y}{y}-\frac{2}{p} d p=0 \Rightarrow \int \frac{d y}{y}-2 \int \frac{d p}{p}=-\log c\)

⇒ \(\log y-2 \log p=-\log c \Rightarrow \log y+\log c=\log p^2\)

⇒ \(\log (c y)=\log p^2 \Rightarrow p^2=y c\) ………………………(4)

Eliminating p from (1) and (4): general solution of (1) is

8 \(y^2=p\left(4 x y-p^2\right)=p(4 x y-c y) \Rightarrow 64 y^4=p^2 y^2(4 x-c)^2\)

⇒ \(64 y^4=c y^3(4 x-c)^2 \Rightarrow 64 y=c(4 x-c)^2\)

Problems On First-Order Non-First-Degree Differential Equations

Example. 6. Solve \(2 p x=2 \tan y+p^3 \cos ^2 y\)

Solution.

Given equation is \(2 p x=2 \tan y+p^3 \cos ^2 y\) ……………………..(1)

Since x is of first degree in (1), it can be solved for x:

∴ \(x=\frac{\tan y}{p}+\frac{p^2 \cos ^2 y}{2}\) ……………………..(2)

Differentiating (2) w.r.t.y:

⇒ \(\frac{1}{p}=\frac{1}{p} \sec ^2 y-\frac{1}{p^2} \tan y \frac{d p}{d y}+p \cos ^2 y \frac{d p}{d y}-p^2 \sin y \cos y\)

⇒ \(\left[\frac{1}{p} \tan ^2 y-p^2 \sin y \cos y\right]+\left(p \cos ^2 y-\frac{\tan y}{p^2}\right) \frac{d p}{d y}=0\)

⇒ \(-p \tan y\left(p \cos ^2 y-\frac{\tan y}{\dot{p}^2}\right)+\left(p \cos ^2 y-\frac{\tan y}{p^2}\right) \frac{d p}{d y}=0\)

⇒ \(\left(p \cos ^2 y-\frac{\tan y}{p^2}\right)\left(\frac{d p}{d y}-p \tan y\right)=0\)

∴ \(p \cos ^2 y-\frac{\tan y}{p^2}=0\) ………………………(3)

∴ \(\frac{d p}{d y}-p \tan y=0\) ……………………..(4)

(3) is rejected as it gives a singular solution.

Solving (4): \(\frac{d p}{d y}=p \tan y \Rightarrow \frac{d p}{p}=\tan y d y\)

⇒ \(\int \frac{d p}{p}=\int \tan y d y+\log c \Rightarrow \log p=\log \sec y+\log c\)

⇒ \(\log p=\log (c \sec y) \Rightarrow p=c \sec y\) ………………………(5)

Eliminating p from (1) and (5) \(\Rightarrow 2 c x y=2 \tan y+c^3 \sec ^3 y \cos ^2 y \Rightarrow 2 c x=2 \sin y+c^3\)

∴ The general solution of (1) is \(2 c x=2 \sin y+c^3\).

Differential Equations of First Order But Not of First Degree Differential Equations Solvable For Y:

Let f(x, y, p) = 0 be the giiven differential equation ………………..(1)

If (1) cannot be resolved into two rational and linear factors and (1) is of first degree in y , then it can be solved for y .

(1) can be expressed in the form y = F(x,p) …………………….(2)

Differentiation of (2) w.r.t.x. gives an equation of the form \(p=g\left(x, p, \frac{d p}{d x}\right)\) ……………………(3)

Since (3) is an equation in two variables p and x, it can be solved.
The solution of (1) is ψ(x,y,c) = 0 ……………….(4)

Note 1. If it is not possible to eliminate P from (1) and (4), the general solution of (1) is given in the form (1) Φ(x, p, c) = 0, f(x,y,p) = 0 or \(x=f_1(p, c), y=f_2(p, c)\)

This is regarded as a parametric form of the required solution where p is regarded as a parameter.

2. This method is especially useful for equations in which x is absent.

Differential Equations of First Order But Not of First Degree Solved Problems

Example. 1. Solve \(y=2 p x-p^2\)
Solution.

Given equation is \(y=2 p x-p^2\) ……………………(1)

Differentiating (1) w.r.t.x => \(p=2 p+2 x \frac{d p}{d x}-2 p \frac{d p}{d x}\)

⇒ \(2(p-x) \frac{d p}{d x}=p \Rightarrow p \frac{d x}{d p}+2 x=2 p \Rightarrow \frac{d x}{d p}+\frac{2 x}{p}=2\) …………………….(2)

(2) is a linear equation in x.

Then \(\text { I.F. }=\exp \left[\int \frac{2}{p} d p\right]=\exp (2 \log p)=\exp \left(\log p^2\right)=p^2\)

∴ The general solution of (2) is x \((\mathrm{I} . \mathrm{F})=\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d p+\mathrm{c}\)     where Q = 2.

⇒ \(x p^2=\int 2 p^2 d p+c=\frac{2 p^3}{3}+c \Rightarrow x=\frac{2}{3} p+\frac{c}{p^2}\) ………………………(3)

It is not possible to eliminate p from (1) and (3)

∴ The general solution of (1) is given by two equations:

x = \((2 p / 3)+\left(c / p^2\right) \text { and } y=2 p\left(\frac{2 p}{3}+\frac{c}{p^2}\right)-p^2=\frac{p^2}{3}+\frac{2 c}{p}\)

Differential Equations Of First Order But Not First Degree Solvable For y 

Example. 2. Solve \(y=x p^2+p\)
Solution.

Given equation is \(y=x p^2+p\) …………………….(1)

Differentiating w.r.t. x => \(p=p^2+2 x p \frac{d p}{d x}+\frac{d p}{d x}\)

⇒ \(\left(p^2-p\right) \frac{d x}{d p}+2 x p+1=0 \Rightarrow \frac{d x}{d p}+\frac{2}{p-1} x=-\frac{1}{p(p-1)}\) ………………………..(2)

(2) is a linear equation in x where \(\mathrm{P}=\frac{2}{p-1}, \quad \mathrm{Q}=\frac{-1}{p(p-1)}\)

⇒ \(\int \mathrm{P} d p=\int \frac{2}{p-1} d p=2 \log (p-1)=\log (p-1)^2\)

∴ \(\text { I.F. }=e^{\log (p-1)^2}\)

∴ The general solution of (2) is \(x(\mathrm{I} . \mathrm{F})=\int \mathrm{Q}(\mathrm{I} . \mathrm{F}) d p+c \Rightarrow x(p-1)^2=\int \frac{-1}{p(p-1)}(p-1)^2 d p+c\)

⇒ \(x(p-1)^2=c-\int \frac{p-1}{p} d p=c-p+\log p\) ……………………..(3)

It is not possible to eliminate 7 from (1) and (3).

∴ The G.S. of (1) is given by \(x=(c-p+\log p)(p-1)^{-2} \text { and } y=x p^2+p\).

Examples Of Differential Equations Solvable For y

Example. 3. Solve \(y+p x=p^2 x^4\)
Solution.

Given equation is \(y+p x=p^2 x^4\) ………………………(1)

But (1) is of first degree in y and hence (1) can be solved for y.

Differentiating (1) w.r.t. x, we get \(p+p+x \frac{d p}{d x}=2 p x^4 \frac{d p}{d x}+4 p^2 x^3 \Rightarrow 2 p+\left(x-2 p x^4\right) \frac{d p}{d x}-4 p^2 x^3=0\)

⇒ \(x\left(1-2 x^3 p\right) \frac{d p}{d x}+2 p\left(1-2 x^3 p\right)=0 \Rightarrow\left(1-2 x^3 p\right)\left(x \frac{d p}{d x}+2 p\right)=0\)

⇒ \(1-2 x^3 p=0\) ………………………..(2)

x\(\frac{d p}{d x}+2 p=0\) ………………………..(3)

(2) is discarded as it gives a singular solution

Solving(3): \(x \frac{d p}{d x}+2 p=0 \Rightarrow \frac{d p}{p}+\frac{2 d x}{x}=0\)

Integrating: \(\int \frac{d p}{p}+2 \int \frac{d x}{x}=\log c \Rightarrow \log p+2 \log x=\log c\)

⇒ \(\log p+\log x^2=\log c \Rightarrow \log p x^2=\log c \Rightarrow p x^2=c \Rightarrow p=c / x^2\)

Eliminating p from (1) and (4), the general solution of (1) is \(y+(c / x)=\left(c^2 / x^4\right) x^4 \Rightarrow y+(c / x)=c^2\)

Note 1. From (2) : \(p=1 /\left(2 x^3\right)\) ………………….(5)

Eliminating p from (1) and (5)

⇒ \(y+\left(1 / 2 x^2\right)=1 /\left(4 / x^2\right) \Rightarrow y+1 /\left(4 x^2\right)=0 \Rightarrow 4 x^2 y+1=0\). which is called a singular solution since it does not contain an arbitrary constant.

2. While finding the G.S. the factor not involving \(\frac{d p}{d x}\) is discarded.

First-Order Equations Not Of First Degree Solvable For y Examples 

Example. 4. Solve \(y=2 x p+x^2 p^4\)
Solution.

Given equation is \(y=2 x p+x^2 p^4\) ………………………..(1)

Differentiating (1) w.r.t. x, we get: \(\frac{d y}{d x}=2 p+2 x \frac{d p}{d x}+2 x p^4+4 x^2 p^3 \frac{d p}{d x}\)

⇒ \(p=2 p+2 x p^4+2 x\left(1+2 x p^3\right) \frac{d p}{d x}\)

⇒ \(p\left(1+2 x p^3\right)+2 x\left(1+2 x p^3\right) \frac{d p}{d x}=0 \Rightarrow\left(1+2 x p^3\right)\left(p+2 x \frac{d p}{d x}\right)=0\)

⇒ \(1+2 x p^3=0\) …………………….(2)

p\(+2 x \frac{d p}{d x}=0\) ……………………(3)

(2) is discarded as it gives a singular solution.

Solving(3): \(p+2 x \frac{d p}{d x}=0 \Rightarrow \frac{2 d p}{p}+\frac{d x}{x}=0\)

Integrating: \(\Rightarrow 2 \int \frac{d p}{p}+\int \frac{d x}{x}=\log c \Rightarrow 2 \log p+\log x=\log c\)

⇒ \(\log p^2 x=\log c \Rightarrow p^2 x=c \Rightarrow p^2=c / x\) …………………….(4)

Eliminating p from (1) and (4) \(\Rightarrow\left(y-x^2 p^4\right)^2=4 x^2 p^2\)

∴ The general solution of (1) is \(\left[y-\left(x^2 c^2 / x^2\right)\right]^2=4 x^2(c / x) \Rightarrow\left(y-c^2\right)^2=4 c x\).

Examples Of Solvable For y Differential Equations 

Example. 5. Solve \(x p^2-2 y p+x=0\)
Solution.

Given equation is \(x p^2-2 y p+x=0\) ………………………..(1)

Differentiating (1) w.r.t. \(x \Rightarrow p^2+2 p x \frac{d p}{d x}-2 p \frac{d y}{d x}-2 y \frac{d p}{d x}+1=0\)

⇒ \(p^2+2(p x-y) \frac{d p}{d x}-2 p^2+1=0 \Rightarrow\left(1-p^2\right)+2(p x-y) \frac{d p}{d x}=0\) ………………………..(2)

Eliminating y from (1) and (2) : \(\left(1-p^2\right)+2\left(p x-\frac{x p^2+x}{2 p}\right) \frac{d p}{d x}=0\)

⇒ \(\left(1-p^2\right)+\frac{2\left(2 p^2 x-x p^2-x\right)}{2 p} \frac{d p}{d x}=0 \Rightarrow p\left(1-p^2\right)+\left(p^2-1\right) x \frac{d p}{d x}=0\)

⇒ \(\left(1-p^2\right)\left(p-x \frac{d p}{d x}\right)=0 \Rightarrow 1-p^2=0\) …………………….(3)

p\(-x \frac{d p}{d x}=0\) ………………………(4)

(3) is discarded as it does not contain \(\) and it gives the singular solution.

Solving (4) : \(p=x \frac{d p}{d x} \Rightarrow \frac{d x}{x}=\frac{d p}{p}\)

Integrating: \(\int \frac{d p}{p}=\int \frac{d x}{x}+\log c \Rightarrow \log p=\log x+\log c \Rightarrow \log p=\log c x \Rightarrow p=c x\) ……………………(5)

Eliminating p from (1) and (5): G.S.. of(1) is \(c^2 x^3-2 c x y+x=0 \Rightarrow c^2 x^2-2 c y+1=0\).

Solved Examples Of Differential Equations Solvable For y 

Example. 6. Solve \(y+x p \log p=(2+3 \log p) p^3\)
Solution.

Given equation is \(y=-x p \log p+(2+3 \log p) p^3\) ………………………..(1)

Solving for y: Differentiating (1) w.r.t. x

⇒ \(p=-p \log p-x(\log p+1) \frac{d p}{d x}+\frac{3}{p} \cdot p^3 \frac{d p}{d x}+(2+3 \log p) 3 p^2 \frac{d p}{d x}\)

⇒ \(p(1+\log p)=-x(\log p+1) \frac{d p}{d x}+9 p^2(1+\log p) \frac{d p}{d x}\)

⇒ \((1+\log p)\left[\left(-x+9 p^2\right) \frac{d p}{d x}-p\right]=0\) ……………………(2)

⇒ \(1+\log p=0\) …………………..(3)

p = \(\left(-x+9 p^2\right) \frac{d p}{d x}\) …………………..(4)

(3) is discarded as it gives the singular solution.

Solving (4): \(p \frac{d x}{d p}+x=9 p^2 \Rightarrow \frac{d x}{d p}+\frac{1}{p} x=9 p\) ……………………(5)

(5) is a linear equation in x.

∴ I.F = \(\exp \left(\int \frac{1}{p} d p\right)=e^{\log p}=p\)

The general solution of(5) is \(x_p=\int 9 p \cdot p d p+c=3 p^2+c\) ……………………..(6)

Eliminating x from (1) and (6): \(y=-\left(3 p^2+\frac{c}{p}\right) p \log p+(2+3 \log p) p^3\)

⇒ \(y=-3 p^3 \log p-c \log p+2 p^3+3 p^3 \log p=2 p^3-c \log p\) …………………….(7)

It is not possible to eliminate p from (6) and (7).

∴ The general solution of (1) is \(x=3 p^2+c p^{-1} \text { and } y=2 p^3-c \log p\)

Example. 7. Solve \(x-y p=a p^2\)
Solution.

Given equation is \(x-y p=a p^2 \Rightarrow y=\frac{x}{p}-a p\) ………………………..(1)

Differentiating (1) w.r.t x => \(\frac{d y}{d x}=p=\frac{1}{p}-\frac{x}{p^2} \frac{d p}{d x}-a \frac{d p}{d x}\)

⇒ \(\left(a p^2+x\right) \frac{d p}{d x}=p\left(1-p^2\right) \Rightarrow p\left(1-p^2\right) \frac{d x}{d p}-x=a p^2\)

⇒ \(\frac{d x}{d p}-\frac{1}{p\left(1-p^2\right)} x=\frac{a p}{1-p^2}\) is a linear equation in x ………………………….(2)

where P = \(-\frac{1}{p\left(1-p^2\right)}\) and \(\mathrm{Q}=\frac{a p}{1-p^2}\)

⇒ \(\int \mathrm{P} d p=\int \frac{1}{p\left(1-p^2\right)} d p=\int\left[\frac{1}{p}+\frac{1}{2(1-p)}-\frac{1}{2(1+p)}\right] d p\)

= \(\log p-\frac{1}{2} \log (1-p)-\frac{1}{2} \log (1+p)=\log \left(\frac{p}{\sqrt{1-p^2}}\right)\)

∴ \(\text { I.F. }=\exp \left[\int \mathrm{P} d p\right]=\exp \left[-\log \frac{p}{\sqrt{1-p^2}}\right]=\exp \left[\log \frac{\sqrt{1-p^2}}{p}\right]=\frac{\sqrt{1-p^2}}{p}\)

The general solution of (2) is \(\frac{x \sqrt{1-p^2}}{p}=\int \frac{a p}{1-p^2} \cdot \frac{\sqrt{1-p^2}}{p} d p+c\)

⇒ \(\frac{x \sqrt{1-p^2}}{p}=a \int \frac{d p}{\sqrt{1-p^2}}+c=a \sin ^{-1} p+c \Rightarrow x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\) …………………….(3)

It is not possible to eliminate p from (1) and (3)

∴ The general solution of (1) is given by the two equations

y = \(x p^2+p \text { and } x=\frac{p}{\sqrt{1-p^2}}\left(a \sin ^{-1} p+c\right)\)

Example 8. Solve : \(x^2+p^2 x=y p\)
Solution.

Given equation is \(x^2+p^2 x=y p\) ……………………….(1)

Solving for y: \(y=p x+\left(x^2 / p\right)\)

Differentiating w.r.t.x : \(\frac{d y}{d x}=p=p+x \frac{d p}{d x}+\frac{p(2 x)-x^2(d p / d x)}{p^2}\)

⇒ \(x \frac{d p}{d x}+\frac{2 p x-x^2(d p / d x)}{p^2}=0 \Rightarrow p^2 x \frac{d p}{d x}+2 p x-x^2 \frac{d p}{d x}=0\)

⇒ \(x\left(p^2-x\right) \frac{d p}{d x}=-2 p x \Rightarrow\left(p^2-x\right) \frac{d p}{d x}=-2 p \Rightarrow\left(x-p^2\right) \frac{d p}{d x}=2 p \Rightarrow 2 p \frac{d x}{d p}-x=-p^2\)

⇒ \(\frac{d x}{d p}-\frac{1}{2 p} x=-\frac{p}{2}\). This is a linear equation in x. ………………………..(2)

Now I.F. = \(e^{\int-(1 / 2 p) d p}=e^{-(1 / 2) \log p}=e^{\log p^{-1 / 2}}=p^{-1 / 2}\)

∴ The general solution of (2) is \(x \cdot p^{-1 / 2}=\int-\frac{p}{2} \cdot p^{-1 / 2} d p+c\)

⇒ \(\frac{x}{\sqrt{p}}=-\frac{1}{2} \int p^{1 / 2} d p+c \Rightarrow \frac{x}{\sqrt{p}}=-\frac{1}{2} \frac{p^{3 / 2}}{3 / 2}+c=-\frac{1}{3} p^{3 / 2}+c\)

⇒ \(x=c \sqrt{p}-(1 / 3) p^2\) …………………………(3)

Substituting x from (3) in (1):

y = \(p\left(c \sqrt{p}-(1 / 3) p^2\right)+\frac{\left[c \sqrt{p}-(1 / 3) p^2\right]^2}{p}\) …………………………(4)

(3) and (4) together is the solution of (1)

Differential Equations of First Order But Not of First Degree Equations Homogeneous In X and Y

When the given equation is homogeneous in x and y it can be written as \(\mathrm{F}\left(\frac{d y}{d x}, \frac{y}{x}\right)=0\).

It is then possible to solve it for \(p=\frac{d y}{d x}\).

Differentiating with respect to x, we get \(p=f(p)+x f^{\prime}(p) \frac{d p}{d x}\) from which we can separate the variables and write \(\frac{d x}{x}=\frac{f^{\prime}(p) d p}{p-f(p)}\) and solve

Differential Equations of First Order But Not of First Degree Solved Problems

Example. 1. Solve the D.E. \(y^2+x y p-x^2 p^2=0\) which is homogeneous in x and y.
Solution.

Given equation is \(y^2+x y p-x^2 p^2=0\) …………………… (1)

Dividing(1)by \(x^2\), we have \(\frac{y^2}{x^2}+\frac{y}{x} p-p^2=0\)

⇒ \(\frac{y}{x}=\frac{-p \pm \sqrt{p^2+4 p^2}}{2}=\frac{1 \pm \sqrt{5}}{2} p \Rightarrow \frac{y}{x}=c \cdot \frac{d y}{d x}\) where c = \(\frac{-1 \pm \sqrt{5}}{2}\)

⇒ \(\frac{d x}{x}=c \cdot \frac{d y}{y}\)

∴ Integrating \(\log x=c \log y+\log c_1\) (where \(c_1\) is a constant)

⇒ \( \log x=\log \left(c_1 y^c\right) \Rightarrow x=c_1 y^c \Rightarrow x-c_1 y^c=0\)

∴ The required general solution of equation (1) is

⇒ \(\left[x-c_1 y^{\left(\frac{-1+\sqrt{5}}{2}\right)}\right]\left[x-c_1 y^{\left(\frac{-1-\sqrt{5}}{2}\right)}\right]=0\)

∴ \(c=\frac{-1 \pm \sqrt{5}}{2}\)

Differential Equations Of First Order But Not Of First Degree Examples

Example. 2. Solve \(8 y^2-4 x y p-x^2 p^2=0\)
Solution.

Given equation is \(8 y^2-4 x y p-x^2 p^2=0\)

Dividing with \(x^2, 8(y / x)^2-4(y / x) p-p^2=0\)

Solving for (y / x), we get \(\frac{y}{x}=\frac{4 p \pm \sqrt{16 p^2+32 p^2}}{2}=(2 \pm 2 \sqrt{3}) p \Rightarrow \frac{y}{x}=(2 \pm 2 \sqrt{3}) \frac{d y}{d x}\)

Separating the variables, \(\frac{d y}{y}=\frac{1}{2 \pm 2 \sqrt{3}} \cdot \frac{d x}{x}\)

Integrating on both sides,

log y \(=\frac{1}{2(1+\sqrt{3})} \log x+\log c \text { and } \log y=\frac{1}{2(1-\sqrt{3})} \log x+\log c\)

⇒ \(y=c x^{1 / 2(1+\sqrt{3})} \text { and } y=c x^{1 / 2(1-\sqrt{3})}\)

C.S. is \(\left[y-c x^{1 / 2(1+\sqrt{3})}\right]\left[y-c x^{1 / 2(1-\sqrt{3})}\right]=0\)

Solved Problems On Homogeneous Equations In x And y

Example. 3. Solve \(y=y p^2+2 p x\)
Solution.

Given equation is \(y=y p^2+2 p x \text { i.e. } y\left(1-p^2\right)=2 p x . \quad y=\frac{2 p x}{1-p^2}\) ………………………(1)

Differentiating w.r.t. x, \(\frac{2 d p}{p\left(1-p^2\right)}+\frac{d x}{x}=0\)

=> Using partial fractions, \(\left(\frac{2}{p}+\frac{1}{1-p}-\frac{1}{1+p}\right) d p+\frac{d x}{x}=0\)

Integrating, \(x=\frac{c\left(1-p^2\right)}{p^2}\) …………………………..(2)

Eliminating p between (1) and (2) we get, \(y^2=4 x c+4 c^2\) which is the required solution.

Differential Equations of First Order But Not of First Degree Clairaut’s Equation

The differential equation of the form y = xp + f (p) is called Clairaut’s equation.

This equation is solved by considering it as y = f(x,p), solvable for the y type.

Solution of Clairaut’s Equation :

Let the given equation be y = xp + f (p) ……………………..(1)

Differentiating (1) w.r.t. x: \(\Rightarrow \frac{d y}{d x}=x \frac{d p}{d x}+p+f^{\prime}(p) \frac{d p}{d x}\)

⇒ \(p=\left[x+f^{\prime}(p)\right] \frac{d p}{d x}+p \Rightarrow\left[x+f^{\prime}(p)\right] \frac{d p}{d x}=0\)

⇒ \(\frac{d p}{d x}=0\) ……………………(2)

x\(+f^{\prime}(p)=0\) …………………….(3)

Solving (2): \(\frac{d p}{d x}=0 \Rightarrow p=c\) where c is a real number. ………………………..(4)

Eliminating p from (1) and (4) y = cx + f(c).

∴ The general solution of Clairaut’s equation (1) is y = cx + f(c).

Note. Solving the equation (3): \(x+f^{\prime}(p)=0\) ……………………….(5)

By eliminating p from (1) and (5) we get the solution Φ (x,y) = 0.

As this solution does not contain any arbitrary constant this is not the general solution. This solution is called the singular solution of (1).

Differential Equations of First Order But Not of First Degree Working Rule for Solving Clairaut’s Equation :

1. The given equation can be written in the form y = xp + f(p) ………………..(a)

2. In order to find the solution of (a), replace p by c where c is any real number.

3. Putting p = c in (a), we get y = xc + f(c).

∴ The general solution of (a) is y = xc + f{c).

Differential Equations of First Order But Not of First Degree Solved Problems

Example. 1. Solve \((y-x p)(p-1)=p\)
Solution.

Given equation is \((y-x p)(p-1)=p\) …………………….(1)

⇒ \(y(p-1)-x p(p-1)=p \Rightarrow y(p-1)=x p(p-1)+p\)

⇒ \(y=x p+\frac{p}{p-1}\) which is in Clairaut’s form.

The genera] solution of (1) is \(y=x c+\frac{c}{c-1}\) where c is any real number

Example. 2. Solve \(p=\tan (x p-y)\)
Solution:

Given equation is \(p=\tan (x p-y)\)

⇒ x p-y=\text{Tan}^{-1} p \Rightarrow y=x p-\text{Tan}^{-1} p[/latex] which is in Clairaut’s form.

∴ The general solution of (1) is \(y=x c-\text{Tan}^{-1} c\) where c is any real number.

Problems On First-Order But Not First-Degree Equations Homogeneous In x And y

Example. 3. Solve \(y^2-2 p x y+p^2\left(x^2-1\right)=m^2\)
Solution.

Given equation is \(y^2-2 p x y+p^2\left(x^2-1\right)=m^2\) ……………………….(1)

⇒ \(y^2-2 p x y+p^2 x^2=p^2+m^2 \Rightarrow(y-p x)^2=p^2+m^2\)

⇒ \(y-p x=\sqrt{p^2+m^2} \Rightarrow y=x p+\sqrt{p^2+m^2}\) which is in Clairaut s form.

∴ The general solution of (1) is \(y=x c+\sqrt{c^2+m^2}\)

Applications Of Clairaut’s Equation In Differential Equations.

Example. 4. Solve \(\sin p x \cos y=\cos p x \sin y+p\)
Solution.

Given equation is \(\sin p x \cos y-\cos p x \sin y=p\) …………………..(1)

⇒ \(\sin (p x-y)=p \Rightarrow p x-y=\sin ^{-1} p \Rightarrow y=p x-\sin ^{-1} p\) which is in Clairaut’s form.

∴ The general solution of (1) is \(y=c x-\sin ^{-1} c\), c being an arbitrary real number.

Solved Problems On Homogeneous Equations Of First Order Not First Degree

Example. 5. Solve \(p=\log (p x-y)\)
Solution.

Given equation is \(p=\log (p x-y)\) ……………….. (1)

⇒ \(p x-y=e^p \Rightarrow y=p x-e^p\) which is in Clairaut’s form.

∴ The general solution of (1) is \(y=c x-e^c\) where c is any real number.