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Factor the numerator by $2$
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$\frac{dy}{dx}=\frac{2\left(xy^2+2\right)}{2\left(3-x^2y\right)}$
Learn how to solve problems step by step online. Solve the differential equation dy/dx=(2xy^2+4)/(2(3-x^2y)). Factor the numerator by 2. Cancel the fraction's common factor 2. Rewrite the differential equation in the standard form M(x,y)dx+N(x,y)dy=0. The differential equation 3-x^2ydy-\left(xy^2+2\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.