Question

Write the differential equation y'e^x + ye^2x - six = 0 in the standard form

Answer 1

Let us write the differential equation $y'e^x + ye^{2x} - \sin x = 0$ in the standard form.

It follows that $y'e^x + ye^{2x} =\sin x,$ and hence $y' + e^{x}y =e^{-x}\sin x$ is the standard form of the differential equation $y'e^x + ye^{2x} - \sin x = 0.$