logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Ad
0 votes

$y-x\big(\large\frac{dy}{dx}\big)=$$a^2(y^2+x^2\large\frac{dy}{dx})$

Can you answer this question?
 
 

1 Answer

0 votes
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}$$+\log c$
$\qquad\qquad\qquad\;\;=\log \large\frac{ca^2x}{a^2x+1}$
But $ca^2$=Constant
$\Rightarrow ca^2=k$
$\qquad\qquad\qquad\;\;=\log\bigg|\large\frac{kx}{a^2x+1}\bigg|$
Step 4:
On substituting these in equ(1) we get
$\therefore \log \bigg|\large\frac{a^2y-1}{a^2-y}\bigg|=$$\log\bigg|\large\frac{kx}{a^2x+1}\bigg|$
$\Rightarrow \large\frac{a^2y-1}{a^2y}=\frac{kx}{a^2x+1}$
$\Rightarrow (a^2y-1)(a^2x+1)=ka^2xy$
$K=ka^2$
$\Rightarrow (a^2y-1)(a^2x+1)=Kxy$
answered Aug 26, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...