Question

use separation of variables to solve the initial value problem dy/dx=5-x^2/3y; y=-1 when x=0

Answer 1

\[3yy'=5-x^2\]
\[3\int \frac{y'}{y}dx=\int 5-x^2 dx\\
3\ln(y)=5x-\frac{x^3}{3}\] is a start

Answer 2

i did \[3ydy=5-x^2dx\]
\[\int\limits3ydy=\int\limits5-x^2dx \]
\[3y^2/2=5x-x^3/3+c\]