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Let $y(x)$ be a solution of $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$ and $y(0)=-1 .$ Then y(1) s equal to
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The correct answer is:
$\frac{1}{6}$
We have. $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^{2}}\right) y=\frac{4 x^{2}}{1+x^{2}}$
Here. $\mathrm{IF}=e^{\frac{1}{1+x^{2}}}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$
$\therefore \quad y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \times \frac{4 x^{2}}{\left(1+x^{2}\right)} d x+C$
$\Rightarrow y\left(1+x^{2}\right)=\int 4 x^{2} d x+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}-1$
$$
[y(0)=-1]
$$
$\Rightarrow \quad y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}-\frac{1}{1+x^{2}}$
$\therefore \quad y(1)=\frac{4}{6}-\frac{1}{2}=\frac{1}{6}$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^{2}}\right) y=\frac{4 x^{2}}{1+x^{2}}$
Here. $\mathrm{IF}=e^{\frac{1}{1+x^{2}}}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$
$\therefore \quad y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \times \frac{4 x^{2}}{\left(1+x^{2}\right)} d x+C$
$\Rightarrow y\left(1+x^{2}\right)=\int 4 x^{2} d x+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}-1$
$$
[y(0)=-1]
$$
$\Rightarrow \quad y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}-\frac{1}{1+x^{2}}$
$\therefore \quad y(1)=\frac{4}{6}-\frac{1}{2}=\frac{1}{6}$
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