Learn how to solve integrals of exponential functions problems step by step online. Solve the differential equation (x^3-2xyx)dx-(x^2+1)dy=0. The differential equation \left(x^3-2xy+x\right)dx-\left(x^2+1\right)dy=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 \frac{x^{4}}{4}-yx^2+\frac{1}{2}x^2 with respect to y to get.

Solve the differential equation (x^3-2xyx)dx-(x^2+1)dy=0