# $y-x\big(\large\frac{dy}{dx}\big)=$$a^2(y^2+x^2\large\frac{dy}{dx}) ## 1 Answer Toolbox: • If the given differential equation is of the form \large\frac{dy}{dx}=$$f(x,y)$,when the variables are separable,we express if in the form $f(x)dx=g(y) dy$
• Then $\int g(y) dy=\int f(x) dx+C$
Step 1:
$y-x\big(\large\frac{dy}{dx}\big)=$$a^2(y^2+x^2\large\frac{dy}{dx}) On expanding we get, y-x\large\frac{dy}{dx}$$=a^2y^2+a^2x^2\large\frac{dy}{dx}$
On rearranging we get,
$\large\frac{dy}{dx}$$(a^2x^2+x)=a^2y^2-y Separating the variables we get, \large\frac{dy}{a^2-y^2-y}=\frac{dx}{a^2x^2+x} (i.e)\large\frac{dy}{y(a^2y-1)}=\frac{dx}{x(a^2x+1)} Step 2: Integrating on both sides we get, \int \large\frac{dy}{y(a^2y-1)}=$$\int \large\frac{dx}{x(a^2x+1)}$------(1)
This can be integrated by the method of partial fraction
Consider $\int \large\frac{dy}{y(a^2y-1)}$
$\large\frac{dy}{y(a^2y-1)}=\frac{A}{y}+\frac{B}{a^2y-1}$
$\Rightarrow A(a^2y-1)+By=1$
Put $y=\large\frac{1}{a^2}$
$\Rightarrow B=a^2$
Put $y=0$
$\Rightarrow A=-1$
$\therefore\int \large\frac{dy}{y(a^2y-1)}=$$\int -\large\frac{dy}{y}+\int \large\frac{a^2}{a^2(y-\Large\frac{1}{a^2})}$$dy$
$\qquad\qquad\;\;=\int -\large\frac{dy}{y}+\int \large\frac{dy}{y-\Large\frac{1}{a^2}}$
$\qquad\qquad\;\;=-\log |y|+\log\big|y-\large\frac{1}{a^2}\big|$
$\qquad\qquad\;\;=\log\bigg|\large\frac{a^2y-1}{a^2y}\bigg|$
Step 3:
Similarly consider $\int \large\frac{dx}{x(a^2x+1)}$
$\large\frac{1}{x(a^2-1)}=\frac{A}{x}+\frac{B}{a^2x+1}$
$\Rightarrow A(a^2x+1)+Bx=1$
Put $x=\large\frac{-1}{a^2}$
$\Rightarrow B(\large\frac{-1}{a^2})$$=1 \Rightarrow B=-a^2 Put x=0 \Rightarrow A=1 \therefore \large\frac{1}{x(a^2x+1)}=\frac{1}{x}-\frac{a^2}{a^2x+1} Now \int\large\frac{1}{x(a^2x+1)}=\int \frac{dx}{x}-\int \frac{a^2}{a^2(x+\Large\frac{1}{a^2})}$$dx$
$\qquad\qquad\qquad\;\;=\log x-\log( x+\Large\frac{1}{a^2})$$+\log c \qquad\qquad\qquad\;\;=\log \large\frac{a^2x}{a^2x+1}$$+\log c$
$\qquad\qquad\qquad\;\;=\log \large\frac{ca^2x}{a^2x+1}$
But $ca^2$=Constant
$\Rightarrow ca^2=k$
$\qquad\qquad\qquad\;\;=\log\bigg|\large\frac{kx}{a^2x+1}\bigg|$
Step 4:
On substituting these in equ(1) we get