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The differential equation $\left(x^2y^3-e^{3x}\right)dx+\left(x^3y^2+y^2e^{\left(y^3\right)}\right)dy=0$ 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 $f(x,y)=C$
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$\left(x^2y^3-e^{3x}\right)dx+\left(x^3y^2+y^2e^{\left(y^3\right)}\right)dy=0$
Learn how to solve differential equations problems step by step online. Solve the differential equation (x^2y^3-e^(3x))dx+(x^3y^2+y^2e^y^3)dy=0. The differential equation \left(x^2y^3-e^{3x}\right)dx+\left(x^3y^2+y^2e^{\left(y^3\right)}\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^{3}y^3}{3}-\frac{1}{3}e^{3x} with respect to y to get.
