Ask Questions, Get Answers

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

The solution of the differential equation $\large\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{1}{(1+x^2)^2}$ is

\[(A)\;y(1+x^2)=c+\tan^{-1}x \quad (B)\;\frac{y}{1+x^2}=c+\tan^{-1}x \quad(C)\;ylog(1+x^2)=c+\tan^{-1}x \quad (D)\;y(1+x^2)=c+\sin^{-1}x\]

Can you answer this question?

1 Answer

0 votes
  • If a differential equation of the form: $ \large\frac{dy}{dx}$$+Py=Q,$ then its general solution is given by $ye^{\int pdx}=\int Q.e^{\int pdx}+c$
Given $\large\frac{dy}{dx}+\frac{2xy}{(1+x^2)}=\frac{1}{(1+x^2)^2}$
This is of the form$:\large\frac{dy}{dx}$$+Py=Q$
Where $P=\large\frac{2x}{1+x^2}$$\;and \; Q=\large\frac{1}{(1+x^2)^2}$
$\int pdx=\int \large\frac{2x}{1+x^2}$$dx$
Put $(1+x^2)=t$ On differentiating w.r.t x we get, $2xdx=dt$
Hence substituting this we get,
$\large\frac{dt}{t}$$=\log t=\log |1+x^2|$
Hence the integraing factor $ e^{\int pdx}=e^{\log (1+x^2)}$
Hence the required solution is
$y \times (I.F)=\int Q.(I.F)dx+c$
$y(1+x^2)=\int =\large\frac{1}{(1+x^2)^2} $$\times (1+x^2)dx+c$
$y(1+x^2)=\int =\large\frac{1}{1+x^2} $$dx+c$
$\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$ \tan ^{-1} \bigg(\large\frac{x}{a}\bigg)$
On integrating we get,
Hence the correct option is $A$


answered May 22, 2013 by meena.p

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