Ask Questions, Get Answers

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

Solve the differential equation $(x^2-1)\Large \frac{dy}{dx}\normalsize +2xy=\Large \frac{1}{x^2-1}$

Can you answer this question?

1 Answer

0 votes
  • Form $\large\frac{dy}{dx}$$+Py=Q,$ where P and Q are constans or functions of x.
  • We find Integrating factor $I.F=e ^{\int pdx}$
  • Now the solution $ y \times I.F=\int [Q \times I.F]dx +c$
  • $ \int \large\frac{dx}{x^2-a^2}=\frac{1}{2a} $$\log \large\frac{|x-a|}{|x+a|}+c$
Given :$(x^2-1) \large\frac{dy}{dx}+$$2xy=\large\frac{1}{x^2-1}$
This is the linear differential equation of the form $\large\frac{dy}{dx}$$+Py=Q$
Divide the equation throughout by $(x^2-1)$ we get
Here $P=\large\frac{2x}{x^2-1}$ and $Q=\large\frac{1}{(x^2-1)^2}$
The solution for $\large\frac{dy}{dx}$$+Py=Q$ is
$y e^{\large\int pdx}$$=\int Qe^{\large\int pdx}+c$
First let us find $\int pdx=I.F$
Here $\int pdx=\int \large\frac{2x}{x^2-1}$$dx$
Put $x^2-1=t$
differentiating with respect to t we get,
Substituting t and dt we get,
$\int \large\frac{dt}{t}$$ =\log |t|$
Substituting back for t we get,
$\int \large\frac{2x}{x^2-1} $$dx=\log |x^2-1|$
Hence $e^{\large\int pdx}$$=e^{\large\log |x^2-1|}$
But $ e^{\large\log x}=x$
Hence $e^{\large\log|x^2-1|}=x^2-1$
Substituting this in solution we get,
$y(x^2-1)=\int \large\frac{1}{(x^2-1)^2}$$.(x^2-1)dx+c$
$=\int \large\frac{1}{(x^2-1)}$$dx$
This is of the form $ \int \large\frac{dx}{x^2-a^2}=\frac{1}{2a} $$\log \large\frac{|x-a|}{|x+a|}+c$
Here $x=x$ and $a=1$
Therefore $ y(x^2-1)=\large\frac{1}{2}$$\; \log \;\large\frac{|x-1|}{|x+1|}+c$
answered May 7, 2013 by meena.p
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