The equation of the curve passing through the origin and satisfying the differential equation (1+x^2 ) d y d x+2 x y=4 x^2 is

Given differential equation is $$\begin{aligned} &\left(1+x^2\right) \frac{d y}{d x}+2 x y=4 x^2 \\ &\Rightarrow \frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{4 x^2}{1+x^2} \end{aligned}$$​(1+x2)dxdy​+2xy=4x2⇒dxdy​+(1+x22x​)y=1+x24x2​​ This is linear diff. equation $$\mathrm{I} . \mathrm{F}=e^{\int \frac{2 x}{1+x^2} d x}=e^{\log \left(1+x^2\right)}=1+x^2$$I.F=e∫1+x22x​dx=elog(1+x2)=1+x2 Solution is $$\begin{aligned} &y\left(1+x^2\right)=\int \frac{4 x^2}{1+x^2} \times 1+x^2+\mathrm{C} \\ &\Rightarrow y\left(1+x^2\right)=\frac{4 x^3}{3}+\mathrm{C} \\ &\Rightarrow \text { Required curve is } \\ &3 y\left(1+x^2\right)=4 x^3(\because \mathrm{C}=0) \end{aligned}$$​y(1+x2)=∫1+x24x2​×1+x2+C⇒y(1+x2)=34x3​+C⇒ Required curve is 3y(1+x2)=4x3(∵C=0)​