Gregory Emery

2021-12-23

Find a general solution $y\left(x\right)$ to the differential equation ${y}^{\prime }=\frac{y}{x}+2{x}^{2}$

Jimmy Macias

$\frac{dy}{dx}=\frac{y}{x}+2{x}^{2}$
$\frac{dy}{dx}-\frac{y}{x}=2{x}^{2}$
(check linear form of differential equation)

$I\cdot F={e}^{-\int pdx}={e}^{-\int -\frac{1}{x}dx}$
$={e}^{\mathrm{log}x}=x$
$⇒y\left(IF\right)=\int Q\left(IF\right)+C$
$⇒yx=\int 2{x}^{3}dx+C$
$⇒xy=\frac{{x}^{4}}{2}+C$
$⇒y=\frac{{x}^{3}}{2}+\frac{c}{x}$ - is the solution

William Appel

Another solution?

user_27qwe

Yes, here:
Solve the linear equation $\left(dy\left(x\right)\right)/\left(dx\right)=2{x}^{2}+y\left(x\right)/x:$
Subtract $y\left(x\right)/x$ from both sides:
$\left(dy\left(x\right)\right)/\left(dx\right)-y\left(x\right)/x=2{x}^{2}$
Let $\mu \left(x\right)={e}^{\left(}int-1/xdx\right)=1/x$
Multiply both sides by $\mu \left(x\right):$
$\left(\left(dy\left(x\right)\right)/\left(dx\right)\right)/x-y\left(x\right)/{x}^{2}=2x$
Substitute $-1/{x}^{2}=d/\left(dx\right)\left(1/x\right):$
$\left(\left(dy\left(x\right)\right)/\left(dx\right)\right)/x+d/\left(dx\right)\left(1/x\right)y\left(x\right)=2x$
Apply the reverse product rule $f\left(dg\right)/\left(dx\right)+g\left(df\right)/\left(dx\right)=d/\left(dx\right)\left(fg\right)$ to the left-hand side:
$d/\left(dx\right)\left(y\left(x\right)/x\right)=2x$
Integrate both sides with respect to x:
$\int d/\left(dx\right)\left(y\left(x\right)/x\right)dx=\int 2xdx$
Evaluate the integrals:
$y\left(x\right)/x={x}^{2}+{c}_{1}$, where ${c}_{1}$ is an arbitrary constant.
Divide both sides by $\mu \left(x\right)=1/x$
$y\left(x\right)=x\left({x}^{2}+{c}_{1}\right)$