2
$\begingroup$

I can solve the differential equation, which is $y + y^{2}/2 = \ln(x) + C$.

But I cannot solve the IVP because I can't isolate for $y$ and find the value of $C.$

$\endgroup$
3
$\begingroup$

We have (just reproducing your work for completeness, and combining some of the comment ideas): \begin{align*} xyy'+xy'&=1\\ xy'(y+1)&=1\\ (y+1)\,dy&=\frac{dx}{x}\\ \frac{y^2}{2}+y&=\ln(x)+C\\ 0&=\ln(1)+C\\ 0&=C\\ \frac{y^2}{2}+y&=\ln(x)\\ y^2+2y&=2\ln(x)\\ y^2+2y+1&=2\ln(x)+1\\ (y+1)^2&=2\ln(x)+1\\ y+1&=\pm\sqrt{2\ln(x)+1}\\ y&=-1\pm\sqrt{2\ln(x)+1}. \end{align*} But notice that we can't allow the negative square root, because it doesn't actually satisfy the initial condition. Also notice that we threw out the $\ln|x|$ for $\ln(x),$ because we knew the initial condition would be for positive $x.$ So the final solution is $$y=-1+\sqrt{2\ln(x)+1}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.