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

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