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The differential equation $\left(\cos\left(x\right)+\ln\left(y\right)\right)dx+\left(\frac{x}{y}+e^y\right)dy=0$ 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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$\left(\cos\left(x\right)+\ln\left(y\right)\right)dx+\left(\frac{x}{y}+e^y\right)dy=0$
Learn how to solve problems step by step online. Solve the differential equation (cos(x)+ln(y))dx+(x/y+e^y)dy=0. The differential equation \left(\cos\left(x\right)+\ln\left(y\right)\right)dx+\left(\frac{x}{y}+e^y\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 \sin\left(x\right)+x\ln\left(y\right) with respect to y to get.
