Question

Solve the Differential Equation:
\[y^\prime\left(2x+y^2\right)=y\]

Answer 1

\[y^\prime\left(2x+y^2\right)=y\]\[\left(2x+y^2\right)\cdot\text dy=y\cdot\text dx\]\[\frac{\text dx}{\text dy}=\frac {2x}y+y\]\[x^\prime-\frac 2y x=y\]

Answer 2

\[R(y)=e^{\int{-\frac{2}y}\text dy}=e^{-2\ln|y|}=y^{-2}=\frac 1{y^2}\]
\[\left(\frac x{y^2}\right)^\prime=\frac{y}{y^2}\]
\[\frac x{y^2}=\int \frac 1y\text dy\]
\[\frac {x}{y^2}=\ln y^2+c_1\]
\[x=y^2\ln y+c_1y^2\]
\[y^2\ln y -x=cy^2\]

Answer 3

there is a mistake on the third line from the bottom
\(\ln y^2\)**\(\ln y\)

Answer 4

Wow, awesome! :D never saw a differential equation being solved this way.

Answer 5

is there another way to solve this question?

Answer 6

There might be, but I just recently learned basic DE's, so I don't really know of another method :(