Exercise
$\left(2x+3y\right)dx+\left(y+3x\right)dy\:=\:0$
Step-by-step Solution
Learn how to solve problems step by step online. Solve the differential equation (2x+3y)dx+(y+3x)dy=0. The differential equation \left(2x+3y\right)dx+\left(y+3x\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 x^2+3yx with respect to y to get.
Solve the differential equation (2x+3y)dx+(y+3x)dy=0
Final answer to the exercise
$3yx+\frac{1}{2}y^2=C_0-x^2$