# Having trouble classifying and solving the ODE $2yy' = y^2 + t - 1$

I seek to classify the ODE $$2yy' = y^2 + t - 1$$ but I am having trouble rearranging the equation such that it can be written as a separable, linear, homogeneous, Bernoulli, or exact ODE.

My first intuition was that it was exact because it can be written like $$-y^2 - t + 1 + 2yy' = 0$$ so $M = -y^2 - t + 1$ and $N = 2y$. However, $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial t}$ in this case, so the criterion for exactness was not met.

I can't seem to fit it into any other of the aforementioned categories, so I'm unsure where to go. Any help or intuition would be greatly appreciated.

• Please note that it is y' or y^{\prime}, not y\prime. Feb 17, 2016 at 9:16
• Thank you - I was unsure which one was appropriate. I assumed that backslash prime was the preferred syntax just because it said prime. It definitely looked a little funny. Feb 17, 2016 at 9:17
• You can also type y'' for the second derivative, instead of the clumsier y^{\prime\prime}, so this “apostrophe notation” is much handier. Feb 17, 2016 at 9:22

Try using the substitution $y^2=m$

• I'm not sure that I know where this would take me. Would I rewrite the equation in terms of $m$ and $t$, solve the DE, and then substitute $y^2$ back in? Additionally, how did you come up with that specific substitution? Feb 17, 2016 at 9:08
• that would take u to the solution mate..as far as how did i come uo with such a substitution is just by intuition and a bit of practice.Just look at the $y^2$ and the $2yy'$ what does that tell u?? Hope u get it and keep on practicing u'll surely improve.best of luck
– user311790
Feb 17, 2016 at 9:10
• Thank you. I suppose inspection is valid here given the $y^2$ and $2yy\prime$. Feb 17, 2016 at 9:14
• Yes and don't say you don't have the brains you just need to apply that tiny bit..just practice Feb 17, 2016 at 9:16

Let $u=y^2$ so, $u'=2yy$ and by substitution in equation, $u'-u+1-t=0$ which has the solution: $u(t)=c\mathbb e^t-t$.
Actually the differential equation in the type $y'+p(x)y=q(x)y^n$ is called Bernoulli differential equation which can be written as $y'y^{-n}+p(x)y^{1-n}=q(x)$. Now, if we substitute $u=y^{1-n}$, then $u'=(1-n)y'y^{-n}$ and so $u'+(1-n)p(x)u=(1-n)q(x)$. your equation $2yy'=y^2+t−1$ can be written as: $y'-\frac12y=\frac12 (t-1)y^{-1}$

• Ah, I see. So this would be a linear DE of the form $y\prime + p(t)y = f(t)$ with $p(t)=-1$ and $f(t) = t - 1$? I have to ask though, how did you and user312254 come up with this substitution? I don't think I have the brains to come up with such a clever solution by inspection. Feb 17, 2016 at 9:13
• @playitright: I hope you have understood the answer. Feb 19, 2016 at 19:07