Solve the following initial value problems: (1+y^2)dx+(x-e^ ^ -1 y )dy=0, y(0)=0

We have, (1+y^2)dx+(x-e^ ^ -1 y )dy=0 (x-e^ ^ -1 y ) dy dx =-(1+y^2) (1+y^2) dx dy =-(x-e^ ^ -1 y ) dx dy + x 1+y^2 = e^ - ^ -1 y 1+y^2 ...(1) Clearly, it is a linear differential equation of the form dy dx +Px=Q Where P=1 1+y^2 and Q= e^ - ^ -1 y 1+y^2 I.F.=e^ =e^ 1 1+y^2 dy =e^ ^ -1 y Multiplying both sides of (1) by I.F.=e^ ^ -1 y , we get e^ ^ -1 y ( dx dy + x 1+y^2 )=e^ ^ -1 y e^ - ^ -1 y 1+y^2 ^ ^ -1 y ( dx dy + x 1+y^2 )=1 1+y^2 Integrating both sides with respect to y, we get e^ ^ -1 y x= 1 1+y^2 dy+C ^ ^ -1 y = ^ -1 y+C ...(2) Now, y(0)=0 0 ^ 0 =0+C =0 Putting the value of C in (2), we get xe^ ^ -1 y = ^ -1 y+0 ^ ^ -1 y = ^ -1 y Hence, xe^ ^ -1 y = ^ -1 y is the required solution.

Solve the following initial value problems: (1+y^2)dx+(x-e^ ^ -1 …

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Question

Solve the following initial value problems:
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0,\text{ y}(0)=0$

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Answer

We have,
$(1+\text{y}^2)\text{dx}+(\text{x}-\text{e}^{\tan^{-1}\text{y}})\text{dy}=0$
$\Rightarrow(\text{x}-\text{e}^{\tan^{-1}\text{y}})\frac{\text{dy}}{\text{dx}}=-(1+\text{y}^2)$
$\Rightarrow(1+\text{y}^2)\frac{\text{dx}}{\text{dy}}=-(\text{x}-\text{e}^{\tan^{-1}\text{y}})$
$\Rightarrow\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\frac{1}{1+\text{y}^2}$ and $\text{Q}=\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdy}}$
$=\text{e}^{\int\frac{1}{1+\text{y}^2}\text{dy}}$
$=\text{e}^{\tan^{-1}\text{y}}$
Multiplying both sides of (1) by $\text{I.F.}=\text{e}^{\tan^{-1}\text{y}},$ we get
$\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\text{e}^{\tan^{-1}\text{y}}\frac{\text{e}^{-\tan^{-1}\text{y}}}{1+\text{y}^2}$
$\Rightarrow\text{e}^{\tan^{-1}\text{y}}\Big(\frac{\text{dx}}{\text{dy}}+\frac{\text{x}}{1+\text{y}^2}\Big)=\frac{1}{1+\text{y}^2}$
Integrating both sides with respect to y, we get
$\text{e}^{\tan^{-1}\text{y}}\text{x}=\int\frac{1}{1+\text{y}^2}\text{dy}+\text{C}$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+\text{C}\ ...(2)$
Now,
$\text{y}(0)=0$
$\therefore\ 0\times\text{e}^{0}=0+\text{C}$
$\Rightarrow\text{C}=0$
Putting the value of C in (2), we get
$\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}+0$
$\Rightarrow\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$
Hence, $\text{x}\text{e}^{\tan^{-1}\text{y}}=\tan^{-1}\text{y}$ is the required solution.

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