Learn how to solve problems step by step online. Solve the differential equation 2ytdy+y^2dt=0. The differential equation 2yt\cdot dy+y^2dt=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(t,y) with respect to t to get. Now take the partial derivative of y^2t with respect to y to get.

Solve the differential equation 2ytdy+y^2dt=0