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The differential equation $\frac{y}{\sqrt{1-y^2}}dy+\frac{x}{\sqrt{1-x^2}}dx=0$ 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 $f(x,y)=C$
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$\frac{y}{\sqrt{1-y^2}}dy+\frac{x}{\sqrt{1-x^2}}dx=0$
Learn how to solve differential equations problems step by step online. Solve the differential equation y/((1-y^2)^(1/2))dy+x/((1-x^2)^(1/2))dx=0. The differential equation \frac{y}{\sqrt{1-y^2}}dy+\frac{x}{\sqrt{1-x^2}}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. Now take the partial derivative of -\sqrt{1-x^2} with respect to y to get.
