The particular solution of $\frac{d y}{d x}=1+x+y^2+x y^2$, when $y(0)=0$. is

The particular solution of $\frac{d y}{d x}=1+x+y^2+x y^2$, when $y(0)=0$. is
  1. $y=\log \left(1+\frac{x^2}{2}\right)$
  2. $y^3=\log \left(1+\frac{x^2}{2}\right)$
  3. $y^2=\tan \left(1+\frac{x^2}{2}\right)$
  4. $y=\tan \left(x+\frac{x^2}{2}\right)$

Solution

$\begin{aligned} & \frac{d y}{d x}=1+x+y^2+x y^2=(1+x)\left(1+y^2\right) \\ & \Rightarrow \int \frac{d y}{1+y^2}=\int(1+x) d x \\ & \Rightarrow \tan ^{-1} y=x+\frac{x^2}{2}+c \\ & \because y(0)=0 \Rightarrow c=0 \\ & \Rightarrow \tan ^{-1} y=x+\frac{x^2}{2} \\ & \Rightarrow y=\tan \left(x+\frac{x^2}{2}\right)\end{aligned}$

Asked in: MHT CET 2022 (07 Aug Shift 2)

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