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Question
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
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Solution
xy = log y + k
Differentiating w.r.t. x, we get
`x dy/dx+ y (1) = 1/y.dy/dx`
∴ `xy dy/dx+ y ^2 = dy/dx`
∴ `dy/dx- x y dy/dx = y^2`
∴ `(1-xy)dy/dx = y^2`
∴ `y' (1-xy) = y^2`
∴ Given function is a solution of the given differential equation.
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