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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $(1+y^2)+(x-e^{\large\tan^{-1}y})\Large \frac{dy}{dx}$=0

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Toolbox:
  • A linear differential equation of the form $ \large\frac{dx}{dy}$$+Py=Q,$ has the general solution as $xe^{\int pdy}=\int Q. e ^{\int pdy}.dy+c$ Where $e^{\int pdy}$ is the integrating factor (I.F)
  • $ \int \large\frac{dx}{x^2+a^2}=\frac{1}{a} $$\tan ^{-1}\bigg(\large\frac{x}{a}\bigg)+c$
Step 1:
Given $(1+y^2)+(x-e^{\tan ^{-1}y})\large\frac{dy}{dx}=0$
This can be written as
$\large\frac{dx}{dy}=\frac{(x-e^{\Large\tan ^{-1}y})}{1+y^2}$
$\large\frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{\Large\tan ^{-1}y}}{1+y^2}$
Clearly this represents a linear differential equation which is of the form
$\large\frac{dx}{dy}$$+Px=Q$
Here $P= \large\frac{+1}{1+y^2}$ and $ Q=\large\frac{e^{\Large\tan ^{-1}y}}{1+y^2}$
Let us first find the integrating factor I.F which $e^{\int pdy}$
$e^{\large\int pdy}=e^{\Large\int \frac{1}{1+y^2}dy}$
Consider $\int \large\frac{1}{1+y^2}$$dy$
On integrating this we get $\tan ^{-1}y$
Therefore $e^{\large \int pdy}=e^{\large\tan ^{-1}y}$
The solution for the linear differential equation is
Step 2:
$x.e^{\large\int pdy}=\int Q.e^{\large\int pdy}dy+c$
Now substituting the I.F we get
$x.e^{\large\tan ^{-1}y}=\int \large\frac{e^{\Large\tan ^{-1}y}}{1+y^2}.e^{\large\tan ^{-1}y}$$dy+c$
$=\int \large\frac{e^{\Large2\tan ^{-1}y}}{1+y^2}$$dy+c$
Put $\tan ^{-1}y=t,$ on differentiating w.r.t x we get,
$\large\frac{dy}{1+y^2}$$=dt$
Now substituting this we get,
$x.e^{\large\tan ^{-1}y}=\int e^{2t}.dt$
On integrating we get
$x.e^{\large\tan ^{-1}y}=\large\frac{1}{2}\int e^{2t}$$+c$
=>$2x.e^{\large\tan ^{-1}y}= e^{2t}$$+c$
Substituting for t we get,
=>$2x.e^{\large\tan ^{-1}y}= e^{\large2\tan ^{-1}y}$$+c$
Hence the required solution is
$2x.\tan ^{-1}y= e^{\large2\tan ^{-1}y}$$+c$
answered May 9, 2013 by meena.p
 
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