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Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$
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$4y+\sin\left(x\right)-2y\cos\left(x\right)^2dy1\left(2y^2\cos\left(x\right)\sin\left(x\right)+y\cos\left(x\right)\right)dx=0$
Learn how to solve problems step by step online. Solve the differential equation dy/dx=(-(2y^2cos(x)sin(x)+ycos(x)))/(4y+sin(x)-2ycos(x)^2). Rewrite the differential equation in the standard form M(x,y)dx+N(x,y)dy=0. The differential equation 4y+\sin\left(x\right)-2y\cos\left(x\right)^2dy1\left(2y^2\cos\left(x\right)\sin\left(x\right)+y\cos\left(x\right)\right)dx=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.
