JEE MAIN - Mathematics (2024 - 1st February Evening Shift - No. 24)
If $\frac{\mathrm{d} x}{\mathrm{~d} y}=\frac{1+x-y^2}{y}, x(1)=1$, then $5 x(2)$ is equal to __________.
Answer
5
Explanation
$\frac{d x}{d y}-\frac{x}{y}=\frac{1-y^2}{y}$
Integrating factor $=\mathrm{e}^{\int-\frac{1}{y} d y}=\frac{1}{y}$
$\begin{aligned} & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\\\ & \frac{x}{y}=\frac{-1}{y}-y+c \\\\ & x=-1-y^2+c y\end{aligned}$
$\begin{aligned} & x(1)=1 \\\\ & 1=-1-1+c \Rightarrow c=3 \\\\ & x=-1-y^2+3 y \\\\ & 5 x(2)=5(-1-4+6) \\\\ & =5\end{aligned}$
Integrating factor $=\mathrm{e}^{\int-\frac{1}{y} d y}=\frac{1}{y}$
$\begin{aligned} & x \cdot \frac{1}{y}=\int \frac{1-y^2}{y^2} d y \\\\ & \frac{x}{y}=\frac{-1}{y}-y+c \\\\ & x=-1-y^2+c y\end{aligned}$
$\begin{aligned} & x(1)=1 \\\\ & 1=-1-1+c \Rightarrow c=3 \\\\ & x=-1-y^2+3 y \\\\ & 5 x(2)=5(-1-4+6) \\\\ & =5\end{aligned}$
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