Exercise
$\left(2xy^2+ye^x\right)dx+\left(2x^2y+e^x-1\right)dy=0$
Step-by-step Solution
Learn how to solve problems step by step online. Solve the differential equation (2xy^2+ye^x)dx+(2x^2y+e^x+-1)dy=0. The differential equation \left(2xy^2+ye^x\right)dx+\left(2x^2y+e^x-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 y^2x^2+ye^x with respect to y to get.
Solve the differential equation (2xy^2+ye^x)dx+(2x^2y+e^x+-1)dy=0
Final answer to the exercise
$y^2x^2+ye^x-y=C_0$