Question

I seem to be missing a factor of 2 at the end.
Solve the given initial-value problem by finding an appropriate integrating factor.
\[(x)dx+(x^2y+4y)dy=0, y(4)=0\]
\[M(x,y)dx+N(x,y)dy\]
\[M_y=0, N_x=2xy\]
\[\mu(x,y)=\exp(\int\frac{2xy}{x}dy)=e^{y^2}\]
\[(xe^{y^2})dx+(x^2ye^{y^2}+4ye^{y^2})dy=0\]
\[\int xe^{y^2}dx=xe^{y^2}+g(y)\]
\[\int (x^2ye^{y^2}+4ye^{y^2})dy, u=y^2, du=2ydy\]
\[\frac{1}{2}\int (x^2e^u + 4e^u) du=\frac{1}{2}(x^2e^u+4e^u)=\frac{x^2e^{y^2}}{2}+2e^{y^2}+h(x)\]
\[F(x,y)=x^2e^{y^2}+4e^{y^2}=c\]
\[F(4,0)=(4)^2e^{0^2}+4e^{0^2}=20\]
\[F(x,y)=x^2e^{y^2}+4e^{y^2}=20\]

Answer 1

Silly me, \[\int xe^{y^2}dx=\frac{x^2e^{y^2}}{2} +g(y)\]
in this problem.