Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

Prove that \(x^2-y^2\;=\;c(x^2+y^2)^2\)is the general solution of differential equation \((x^3-3x\;y^2)dx\;=\;(y^3-3x^2y)dy\) where \(c\) is a parameter.

Can you answer this question?

1 Answer

0 votes
  • A differential equation of the form $\large\frac{dy}{dx} =$$ F(x,y)$ is said to be homogenous if F(x,y) is a homogenous function of degree zero. To solve such equations, substitute $y = vx$ and $\large\frac{dy}{dx} = v+$$x\large\frac{dv}{dx}.$
Step 1:
Given :$(x^3 - 3xy^2) = (y^3 - 3x^2y)dy$
Rearranging the equation we get,
$\large\frac{dy}{dx} = \frac{(x^3-3xy^2)}{(y^3 - 3x^2y)}$
This is homogenous equation of degree zero.
Hence to solve this let us substitute $y = vx$ and $\large\frac{dy}{dx} = v+$$x\large\frac{dv}{dx}.$
$v+ x\large\frac{dv}{dx} =\frac{ (x^3 -3xv^2x^2)}{(v^3x^3 -3x^2vx)}$
Taking $x^3$ as the common factor and cancelling we get
$v+x\large\frac{dv}{dx} =\frac{ (1-3v^2)}{(v^3 - 3v)}$
Bringing $v$ from the LHS to the RHS we get,
$x\large\frac{dv}{dx }=\frac{ (1-3v^2)}{v(v^2-3)}$$ - v$
$x\large\frac{dv}{dx }= \frac{(1-3v^2 -v^4 + 3v^2)}{v(v^2 - 3)}$
$x\large\frac{dv}{dx }= \frac{(1-v^4)}{v(v^2-3)}$
Seperating the variables we get,
$\large\frac{(v^3-3v)dv}{(1-v^4) }=\frac{ dx}{x}$
$\large\frac{v^3}{(1-v^4)} - \frac{3v}{(1-v^4)} = \frac{dx}{x}$
Step 2:
Integrating on both sides we get,
$\int\large\frac{v^3}{(1-v^4)} -\int \frac{3v}{(1-v^4)} =\int \frac{dx}{x}$
$\int\large\frac{ v^3}{(1-v^4) } = I_1$
This can be done by substitution method
Let $1-v^4 = t$; then $dt = - 3v^3dv$ or $dv = - dt/3$
$I_1 = - (1/3)\int dt/t =- (1/3)\log t$
Substituting for t we get
$I_1 = - (1/3)\log(1-v^4)$
Step 3:
Let $I_2$ be $\large\frac{3v}{(1-v^4)}$
This can be written as
$I_2 = 3 \int \large\frac{v}{(1-v^4)}$
Let $v^2=u$
Hence $du=2vdv$
Or $dv=\large\frac{du}{2}$
Substituting this in $I_2$ we get,
$I_2=\large\frac{1}{2}$$\times 2\log[\large\frac{|1-u|}{|1+u|}]$
$\quad=3/4 \log\large\frac{|1-v^2|}{|1+v^2|}$
$I_1 - I_2 = \large\frac{-1}{4}$$(\log(1-v^4) +\large\frac{3}{4}$$\log\large\frac{|1+v^2|}{|1-v^2|}$
Step 4:
Substituting for $v =\large\frac{ y}{x}$
$-\large\frac{1}{4}$$\log(1-y^4/x^4) +\large\frac{ 3}{4}$$\log\large\frac{(1+y^2/x^2)}{1-(y^2/x^2)}$$ \log x +\log C $
$\log[(x^4 - y^4). (x^2+y^2)^3/x^2-y^2)^3] = 4 \log Cx$
$(x^2+y^2) (x^2+y^2)^3]/(x^2-y^2)^3 = x^4C$
$x^4(x^2+y^2)^4/(x^2-y^2) =C x^4$
$C (x^2 + y^2)^2 = (x^2 - y^2)$
Hence proved.
answered Jul 30, 2013 by sreemathi.v

Related questions

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