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The differential equation $\sec\left(x\right)^2\tan\left(y\right)\cdot dx+\sec\left(y\right)^2\tan\left(x\right)\cdot dy=0$ 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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$\sec\left(x\right)^2\tan\left(y\right)\cdot dx+\sec\left(y\right)^2\tan\left(x\right)\cdot dy=0$
Learn how to solve differential equations problems step by step online. Solve the differential equation sec(x)^2tan(y)dx+sec(y)^2tan(x)dy=0. The differential equation \sec\left(x\right)^2\tan\left(y\right)\cdot dx+\sec\left(y\right)^2\tan\left(x\right)\cdot 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 \tan\left(y\right)\tan\left(x\right) with respect to y to get.
