I'm doing self study and I can't solve this equation:

$$ax + \ln y = y + b$$

Where I'm supposed to eliminate the arbitrary constants.

The given answer is $(y - y^2)(y'') = (y')^2$

But my workings got me to $\left( \frac { 1 }{ \ln { y } } -1 \right) ({ y }^{ \prime \prime })=(y')^{ 2 }$

Could someone provide me a stepwise solution so I know where I went wrong?

Eternally grateful, Confused


Differentiating both sides of the equation, using that $(\ln(y))'=\frac{y'}{y}$ gives $$a+\frac{y'}{y}=y'$$ doing it again leads to $$\frac{y''y-y'^2}{y^2}=y''$$ Which then can be rewritten as $$y''y-y'^2=y''y^2\iff y'^2=y''(y-y^2)$$

  • $\begingroup$ You are right. Thanks for spotting my mistake $\endgroup$ – b00n heT Jul 25 '16 at 13:17

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