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The solution of the differential equation $(1+y^{2})+(x-e^{\tan^{ -1}y})\frac{dy}{dx}=0$ is

Option 1)
$2xe^{\tan^{ -1}\, y}=e^{2\tan^{-1} \, y}+k$

Option 2)
$xe^{\tan^{-1} \, y}=\tan ^{ -1}y+k$

Option 3)
$xe^{2\tan^{-1} \, y}=e^{\tan^{-1} \, y}+k$

Option 4)
$(x-2)=ke^{-\tan ^{-1}y}$

Answers (1)

As we learnt in

Bernoulli's Equation -

$\frac{1}{y^{n-1}}= v$

$\frac{1}{y^{n}}\frac{dy}{dx}= \frac{1}{\left ( 1-n \right )}\frac{dv}{dx}$

- wherein

$\frac{1}{y^{n}}\frac{dy}{dx}+\frac{p}{y^{n-1}}=Q$

$(1+y^{2})+(x-e^{+tan^{-1}y})\frac{dy}{dx}=0$

$=>divided\ by (1+y^{2})$

$1+(\frac{1}{1+y^{2}}\frac{dy}{dx}) (x-e^{+tan^{-1}y})=0$

Now put $tan^{-1}y=t$

$\frac{1}{1+y^{2}}\frac{dy}{dx}=\frac{dt}{dx}$

$=>1+\frac{dt}{dx}(x-e^{t})=0$

$\Rightarrow (x-e^{t})=-\frac{dx}{dt}$

$=> e^{t}-x=\frac{dx}{dt}$

$=>\frac{dx}{dt}+x=e^{t}$

$P=1, Q =e^{t}$

$\int 1.dt=t$

$\therefore I.F.=e^{t}$

$x.e^{t}=\int e^{t}\times e^{t}dt=\frac{1}{2}e^{2t}+k$

$x.e^{tan^{-1}y}=\frac{1}{2}.e^{+2tan^{-1}y}+k$

$=>2xe^{tan^{-1}y}=e^{2tan^{-1}y}+k$

Option 1)

$2xe^{\tan^{ -1}\, y}=e^{2\tan^{-1} \, y}+k$

This option is correct

Option 2)

$xe^{\tan^{-1} \, y}=\tan ^{ -1}y+k$

This option is incorrect

Option 3)

$xe^{2\tan^{-1} \, y}=e^{\tan^{-1} \, y}+k$

This option is incorrect

Option 4)

$(x-2)=ke^{-\tan ^{-1}y}$

This option is incorrect

Posted by

Sabhrant Ambastha

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