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