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The solution of $\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$ is :
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Verified Answer
The correct answer is:
$3 y\left(1+x^2\right)=4 x^3+c$
$\left(1+x^2\right) \frac{d y}{d x}+2 x y-4 x^2=0$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}$
On comparing with $\frac{d y}{d x}+P y=Q$, we get
$P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2}$
I.F. $=e^{\int P d x}=e^{\log \left(1+x^2\right)}$
$=1+x^2$
The solution is
$y \text { (I.F.) }=\int Q \text { (I. F. ) } d x+c_1$
$\Rightarrow \quad y\left(1+x^2\right)=\int \frac{4 x^2}{\left(1+x^2\right)}\left(1+x^2\right) d x+C_1$
$\Rightarrow \quad y\left(1+x^2\right)=\frac{4 x^3}{3}+c_1$
$\Rightarrow \quad 3 y\left(1+x^2\right)=4 x^3+c$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2}$
On comparing with $\frac{d y}{d x}+P y=Q$, we get
$P=\frac{2 x}{1+x^2}, Q=\frac{4 x^2}{1+x^2}$
I.F. $=e^{\int P d x}=e^{\log \left(1+x^2\right)}$
$=1+x^2$
The solution is
$y \text { (I.F.) }=\int Q \text { (I. F. ) } d x+c_1$
$\Rightarrow \quad y\left(1+x^2\right)=\int \frac{4 x^2}{\left(1+x^2\right)}\left(1+x^2\right) d x+C_1$
$\Rightarrow \quad y\left(1+x^2\right)=\frac{4 x^3}{3}+c_1$
$\Rightarrow \quad 3 y\left(1+x^2\right)=4 x^3+c$
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