The equation of the curve passing through the origin and satisfying the differential equation…
The equation of the curve passing through the origin and satisfying the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2$ is
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$\left(1+x^2\right) y=x^3$
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$3\left(1+x^2\right) y=2 x^3$
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$\left(1+x^2\right) y=3 x^3$
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$3\left(1+x^2\right) y=4 x^3$
Solution
Given differential equation is
$
\begin{aligned}
&\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2 \\
&\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}
\end{aligned}
$
This is linear diff. equation
$
\mathrm{I} . \mathrm{F}=e^{\int \frac{2 x}{1+x^2} d x}=e^{\log \left(1+x^2\right)}=1+x^2
$
Solution is
$
\begin{aligned}
&y\left(1+x^2\right)=\int \frac{4 x^2}{1+x^2} \times 1+x^2+\mathrm{C} \\
&\Rightarrow y\left(1+x^2\right)=\frac{4 x^3}{3}+\mathrm{C} \\
&\Rightarrow \text { Required curve is } \\
&3 y\left(1+x^2\right)=4 x^3(\because \mathrm{C}=0)
\end{aligned}
$
Asked in: JEE Main 2013 (25 Apr Online)
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