Differential Equations of First Order But Not of First Degree Equations Reducible To Clairaut’s Form
Some differential equations can be transformed to Clairaut’s form by suitable substitution.
For example : \(y^2=p x y+f\left(\frac{p y}{x}\right)\) ………………….(1)
Put \(x^2=X\) and \(y^2=\mathrm{Y} \Rightarrow 2 x d x=d \mathrm{X}\) and \(2 y d y=d \mathbf{Y}\)
Let \(\mathrm{P}=\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{y}{x} \frac{d y}{d x}=\frac{y}{x} p\) …………………….(2)
(1) and (2) \(\Rightarrow \mathrm{Y}=x y\left(\frac{x \mathrm{P}}{y}\right)+f(\mathrm{P})=x^2 \mathrm{P}+f(\mathrm{P})\) …………………..(3)
=> Y = PX+f(P) which is Clairaut’s differential equation.
∴ The general solution of (3) is Y = cX+f(c) where c is any real number.
⇒ \(y^2=c x^2+f(c)\) is the general solution of (1).
Differential Equations of First Order But Not of First Degree Solved Problems
Example. 1. Solve \(x^2(y-p x)=p^2 y\)
Solution.
Given equation is \(x^2(y-p x)=y p^2\) ……………………..(1)
Put \(x^2=\mathrm{X} \text { and } y^2=\mathrm{Y} \Rightarrow 2 x d x=d \mathrm{X} \text { and } 2 y d y=d \mathrm{Y}\)
⇒ \(\mathrm{P}=\frac{d \mathrm{y}}{d \mathrm{x}}=\frac{2 y d y}{2 x d x}=\frac{y}{x} \frac{d y}{d x}=\frac{y}{x} p \Rightarrow p=\frac{\mathrm{P} x}{y}\) ……………………….(2)
(1) and (2) \(\Rightarrow \mathrm{X}\left(y-x \cdot \frac{\mathrm{P} x}{y}\right)=y \cdot \frac{\mathrm{P}^2 x^2}{y^2} \Rightarrow \mathrm{X}\left(y^2-x^2 \mathrm{P}\right)=x^2 \mathrm{P}^2\)
⇒ \(X(Y-X P)=X P^2 \Rightarrow Y-X P=P^2 \Rightarrow Y=X P+P^2\) ……………………(3)
(3) is Clairaut’s equation.
∴ General solution of (3) is \(Y=C X+C^2\)
∴ The general solution of (1) is \(y^2=C x^2+C^2\)
Differential Equations Of First Order But Not First Degree Reducible To Clairaut’s Form
Example. 2. Solve \((p y+x)(p x-y)=2 p\)
Solution.
Given equation is \((p y+x)(p x-y)=2 p\) ………………………..(1)
Put \(x^2=\mathrm{X} \text { and } y^2=\mathrm{Y} \Rightarrow 2 x d x=d \mathrm{X}, 2 y d y=d \mathrm{Y}\)
⇒ \(\mathrm{P}=\frac{d \mathrm{Y}}{d \mathrm{X}}=\frac{y}{x} \frac{d y}{d x}=\frac{y}{x} \cdot p \Rightarrow p=\frac{x}{y} \mathrm{P}\) …………………………..(2)
(1) and (2) \(\Rightarrow\left(\frac{x}{y} \mathrm{P} \cdot y+x\right)\left(\frac{x}{y} \mathrm{P}, x-y\right)=2\left(\frac{x}{y}\right) \mathrm{P}\)
⇒ \(x(\mathrm{P}+1)\left(x^2 \mathrm{P}-y^2\right)=2 x \mathrm{P} \Rightarrow(\mathrm{P}+1)\left(x^2 \mathrm{P}-y^2\right)=2 \mathrm{P}\)
⇒ \((\mathrm{P}+1)(\mathrm{XP}-\mathrm{Y})=2 \mathrm{P} \Rightarrow \mathrm{PX}-\mathrm{Y}=\frac{2 \mathrm{P}}{\mathrm{P}+1}\)
⇒ \(\mathrm{Y}=\mathrm{PX}-\frac{2 \mathrm{P}}{\mathrm{P}+1}\) is Clairaut’s equation. ………………………….(3)
The general solution of (3) is \(Y=c X-\frac{2 c}{c+1}\)
∴ The general solution of (1) is \(y^2=c x^2-\frac{2 c}{c+1}\)
Solved Problems On First-Order Non-First-Degree Reducible To Clairaut’s Form
Example. 3. Use the transformation \(u=x^2\) and \(v=y^2\) to solve \(a x y p^2+\left(x^2-a y^2-b\right) p-x y=0\).
Solution.
Given: \(a x y p^2+\left(x^2-a y^2-b\right) p-x y=0\).
Put \(x^2=u\) and \(y^2=v \Rightarrow 2 x dx=du, 2y dy=d v\)
⇒ \(\mathrm{P}=\frac{d v}{d x}=\frac{y}{x} \frac{d y}{d x}=\frac{y}{x} p \Rightarrow p=\frac{x}{y} \mathrm{P}\)
(1) and (2) ⇒ \(a x y\left(\frac{x^2}{y^2}\right) \mathrm{P}^2+\left(x^2-a y^2-b\right) \frac{x}{y} \mathrm{P}-x y=0\)
⇒ \(a x^2 \mathrm{P}^2+\left(x^2-a y^2-b\right) \mathrm{P}-y^2=0 \Rightarrow a u \mathrm{P}^2+(u-a v-b) \mathrm{P}-v=0\)
⇒ \(a u \mathrm{P}^2+u \mathrm{P}-a v \mathrm{P}-b \mathrm{P}-v=0 \Rightarrow u \mathrm{P}(a \mathrm{P}+1)-v(a \mathrm{P}+1)-b \mathrm{P}=0\)
⇒ \(v(a \mathrm{P}+1)=u \mathrm{P}(a \mathrm{P}+1)-b \mathrm{P} \Rightarrow v=u \mathrm{P}-\frac{b \mathrm{P}}{a \mathrm{P}+1} \ldots \ldots\) (3) is in Clairaut’s form.
The general solution of (3) is \(v=u \mathrm{C}-\frac{b \mathrm{C}}{a \mathrm{C}+1}\)
∴ The general solution of (1) is \(y^2=x^2 \mathrm{C}-\frac{b \mathrm{C}}{a \mathrm{C}+1}\)
Problems On Equations Reducible To Clairaut’s Form
Example. 4. Reduce the equation \(y^2(y-x p)=x^4 p^2\) to Clairaut’s form by the substitution \(x=1 / u, y=1 / v\) and hence solve the equation.
Solution.
Given equation is \(y^2(y-x p)=x^4 p^2\) ………………………(1)
Also given \(x=1 / u \text { and } y=1 / v \Rightarrow d x=-\frac{1}{u^2} d u, d y=-\frac{1}{v^2} d v\)
⇒ \(\frac{d y}{d x}=\frac{u^2}{v^2} \frac{d v}{d u} \Rightarrow p=\frac{u^2}{v^2} \mathrm{P}\) where \(\mathrm{P}=\frac{d v}{d u}, p=\frac{d y}{d x}\) ……………………..(2)
(1) and (2) \(\Rightarrow \frac{1}{v^2} \cdot\left(\frac{1}{v}-\frac{1}{u} \cdot \frac{u^2}{v^2} \mathrm{P}\right)=\frac{1}{u^4} \cdot \frac{u^4}{v^4} \mathrm{P}^2\)
⇒ \(\frac{1}{v^4}(v-u p)=\frac{\mathrm{P}^2}{v^4} \Rightarrow v-u \mathrm{P}=\mathrm{P}^2 \Rightarrow v=u \mathrm{P}+\mathrm{P}^2\) ……………………..(3)
The general solution of (3) is \(v=u c+c^2\)
Hence the general solution of (1) is \(\frac{1}{y}=\frac{1}{x} c+c^2 \Rightarrow x=c y+c^2 y x\)