Question

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

Answer 1

Answer: 12

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