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

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

1 Answer

  • 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

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