Question

Let the solution curve $y=y(x)$ of the differential equation $\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x=0$ pass through the origin. Then $y (2)$ is equal to _________.

Answer 1

$\left(4+x^{2}\right) d y-2 x\left(x^{2}+3 y+4\right) d x$
$\left(x^{2}+4\right) \frac{d y}{d x}=2 x^{3}+6 x y+8 x$
$\left(x^{2}+4\right) \frac{d y}{d x}-6 x y=2 x^{3}+8 x$
$\frac{d y}{d x}-\frac{6 x}{x^{2}+4} y=\frac{2 x^{3}+8 x}{x^{2}+y}$
L.I. $\frac{ dy }{ dx }+ py =\phi$
$p =\frac{-6 x }{ x ^{2}+4} \phi=\frac{2 x ^{3}+8 x }{ x ^{2}+4}$
$\text { I.F. }=e^{-\int \frac{6 x}{x^{2}+4} d x}=e^{-3 \log _{e}\left(x^{2}+4\right)} $
$=e^{\log _{e}\left(x^{2}+4\right)^{-3}}=\frac{1}{\left(x^{2}+4\right)^{3}}$
Sol
$y \cdot \frac{1}{\left( x ^{2}+4\right)^{3}}=\int \frac{2 x ^{3}+8 x }{\left( x ^{2}+4\right)^{3}\left( x ^{2}+4\right)} dx$
$\frac{ y }{\left( x ^{2}+4\right)^{3}}=\int \frac{2 x \left( x ^{2}+4\right)}{\left( x ^{2}+4\right)^{3}\left( x ^{2}+4\right)} dx$
$x ^{2}+4= t$
$2 xdx = dt$
$\frac{ y }{\left( x ^{2}+4\right)^{3}}=\int \frac{ dt }{ t ^{3}}$
$\frac{ y }{\left( x ^{2}+4\right)^{3}}=\frac{-1}{2\left( x ^{2}+4\right)^{2}}+ C$
passes through origin $(0,0)$
$0=\frac{-1}{2 \times 16}+ C$
$\frac{ y }{\left( x ^{2}+4\right)^{3}}=\frac{-1}{2\left( x ^{2}+4\right)^{2}}+\frac{1}{32}$
$y =\frac{-\left( x ^{2}+4\right)}{2}+\frac{\left( x ^{2}+4\right)^{3}}{32}$
$y (2)=-\frac{8}{2}+\frac{8 \times 8 \times 8}{32}=12$