How do you derive the general solution of y y'' - (y')^2 + y' = 0 Thanks.
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Never mind. Solved.
Let u = y'. Then y'' = u' = \frac{du}{dy} . y' = \frac{du}{dy} . u
Hence the equation is $ yu \cdot\frac{du}{dy} = u^2 - u $ which is separable.
How do you derive the general solution of y y'' - (y')^2 + y' = 0 Thanks.
======
Never mind. Solved.
Let u = y'. Then y'' = u' = \frac{du}{dy} . y' = \frac{du}{dy} . u
Hence the equation is $ yu \cdot\frac{du}{dy} = u^2 - u $ which is separable.
Hint: Divide everything by $y^2$. Then use the identities \left(\dfrac{y'}y\right)'=\dfrac{y''y-(y')^2}{y^2}\qquad\text{and}\qquad\left(\dfrac1y\right)'=\dfrac{-y'}{y^2}.