Learn how to solve differential equations problems step by step online. Solve the differential equation x^2dy+(2xy-e^x)dx=0. The differential equation x^2dy+\left(2xy-e^x\right)dx=0 is exact, since it is written in the standard form M(x,y)dx+N(x,y)dy=0, where M(x,y) and N(x,y) are the partial derivatives of a two-variable function f(x,y) and they satisfy the test for exactness: \displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form f(x,y)=C. Using the test for exactness, we check that the differential equation is exact. Integrate M(x,y) with respect to x to get. Now take the partial derivative of yx^2-e^x with respect to y to get.

Solve the differential equation x^2dy+(2xy-e^x)dx=0