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JEE MAIN 2024
1-2-2024 S2
Question
If $\frac{d x}{d y}=\frac{1+x-y^2}{y}, x(1)=1$, then $5 x(2)$ is equal to :
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Solution
$\begin{aligned} & \frac{d x}{d y}-\frac{x}{y}=\frac{1-y^2}{y} \\ & \text { Integrating factor }=e^{\int-\frac{1}{y} d y}=\frac{1}{y} \\ & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\ & \frac{x}{y}=\frac{-1}{y}-y+c \\ & x=-1-y^2+c y \\ & x(1)=1\end{aligned}$
$\begin{aligned} & 1=-1-1+c \Rightarrow c=3 \\ & x=-1-y^2+3 y \\ & 5 x(2)=5(-1-4+6) \\ & =5\end{aligned}$
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