Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation: xy = log y+C : y'= \(\frac{y^2}{1-xy}\,(xy\neq 1)\)
Solution and Explanation
xy= log y+C
Differentiating both sides of this equation with respect to x, we get:
\(\frac{d}{dx}(xy)=\frac{d}{dx}\,(log \,y)\)
\(\Rightarrow y.\frac{d}{dx}(x)+x.\frac{dy}{dx}=\frac{1}{y}\frac{dy}{dx}\)
\(\Rightarrow y+xy'=\frac{1}{yy'}\)
\(\Rightarrow y^2+xyy'=y'\)
\(\Rightarrow y'=\frac{y^2}{1-xy}\)
∴ L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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Concepts Used:
General Solutions to Differential Equations
A relation between involved variables, which satisfy the given differential equation is called its solution. The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution and the solution free from arbitrary constants is called particular solution.
For example,
Read More: Formation of a Differential Equation