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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2,\text{y}(0)=1$
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Answer

$\frac{\text{dy}}{\text{dx}}=1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2,\text{y}(0)=1$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=(1+\text{x}^2)(1+\text{y}^2)$
$\Rightarrow\frac{\text{dy}}{(1+\text{y}^2)}=(1+\text{x}^2)\text{dx}$
Integrating both sides, we get
$\int \frac{\text{dy}}{(1+\text{y}^2)}=\int(1+\text{x}^2)\text{dx}$
$\Rightarrow\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\text{C}...(1)$
We know that at $\text{x}=0,\text{y}=1.$
Substituting the values of x and y in (1), we get
$\frac{\pi}{4}=0+0+\text{C}$
$\Rightarrow\text{C}=\frac{\pi}{4}$
Substituting the value of C in (1), we get
$\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\frac{\pi}{4}$
Hence, $\tan^{-1}\text{y = x}+\frac{\text{x}^3}{3}+\frac{\pi}{4}$ is the required solution.

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