xy^3dx + (y+1)e^-xdy

Question

Differential Equation (Variable Separation)

Answer 1

This is a differential expression, not an equation, so I assume that the expression equals zero.

xy³dx+(y+1)e⁻ˣdy=0

Rewriting:

(y+1)e⁻ˣdy=-xy³dx

-((y+1)/y³)(dy)=xeˣdx

(-1/y²-1/y³)dy=xeˣdx

Integrating each side:

1/y+1/(2y²)=xeˣ-eˣ+C where C is the integration constant.

This can be written:

2y+1=2xy²eˣ-2y²eˣ+2Cy² or 2y²(xeˣ-eˣ+C).

It can also be written:

2y²(xeˣ-eˣ+C)-2y-1=0.

If the differential expression was a constant a (which may or may not be zero), then the solution would be:

2y²(xeˣ-eˣ+ax+C)-2y-1=0.