Ask Questions, Get Answers

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

Find the equation of a curve passing through origin and satisfying the differential equation $(1-x^2)\Large \frac{dy}{dx}\normalsize-2xy=4x^2$

Can you answer this question?

1 Answer

0 votes
  • If the given linear differential equation is of the form $ \large\frac{dy}{dx}$$+Py=Q,$ then its general solution is $ye^{\int pdx}=\int Q e ^{\int pdx}.dx+c$
  • $\int\large\frac{dx}{x+a}$$=\log |x+a|+c$
Step 1:
Given equation of the curve $(1-x^2)\large\frac{dy}{dx}$$-2xy=4x^2$
Divide throughout by $(1-x^2)$
Clearly this is a linear differential equation of the form
Where $P=-\large\frac{2x}{1-x^2}$ and $ Q=\large\frac{4x^2}{1-x^2}$
Let us now find the integrating factor I.F
$e^{\int pdx}=e^{\int \Large\frac{-2x}{1-x^2}}dx$
Step 2:
Consider $\int \large\frac{-2x}{1-x^2}$$dx$
Put $ 1-x^2=t$ on differentiating w.r.t t
We get $-2xdx=dt$
Now substituting this we get,
$\large\int \frac{dt}{t}$$=\log |t|$
Substituting for t we get,
$\log |1-x^2|$
Hence I.F is $e^{\large\log(1-x^2)}$
but $e^{\large\log x}$$=x$
Hence $I.F=1-x^2$
Now the required solution is
$ye^{\large\int pdx}$$=\int Q. e ^{\large\int pdx}$$dx+c$
$y.(1-x^2)=\int \large\frac{4x^2}{1-x^2}$$(1-x^2)dx+c$
=>$y(1-x^2)=\int 4x^2.dx+c$
On integrating we get
Step 3:
It is given that the curve passes through the orgin, which means $x=0,y=0$
Let us now substitute for x and y to evaluate the value for c
$0(1+0)=4 \times 0+c=>c=0$
Hence the required solution is
answered May 9, 2013 by meena.p

Related questions

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