Question

what is the gen. solution for this D.E :xy^3 dx + e^x^2 dy?

Answer 1

I'm assuming this d.e. is set equal to zero. This is separable. \[xy^3dx+e^{x^2}dy=0 \rightarrow xy^3dx=-e^{x^2}dy \rightarrow \frac{x}{e^{x^2}}dx=-\frac{1}{y^3}dy\]To solve the right-hand side, you can make the substitution \[u=x^2\]for then\[du=2x dx \rightarrow \frac{1}{2}du=x dx\]You can write, then,\[\frac{x dx}{e^{x^2}}= \frac{1}{2}\frac{du}{e^u}\]

Answer 2

Integrating both sides,\[\frac{1}{2}(-e^{-u})=\frac{1}{2}y^{-2}\]

Answer 3

\[-e^{-x^2}=\frac{1}{y^2}+c \rightarrow (c+e^{-x^2})y^2+1=0\]