In a practice exam, we're asked to solve the differential equation $y'=(y^2-1)e^{ty}$, with $y(1)=0$.

However, the only techniques we've covered are separable differential equations, first order linear differential equations with integration factors or variation of parameters, and differential forms.

This looks like a nonlinear differential equation because I can't separate the $ty$ in the exponent. I wrote it as a form $$ (1-y^2)e^{ty}dt+1dy=0 $$ but this is not exact, and I can't find an integration factor, since the usual formulas don't end up being expressions only in $t$ or $y$. It's not homogeneous either. I think it might be a typo since it's not something we covered, but is there a way to solve it I'm overlooking?

  • $\begingroup$ I wonder if there is a typo? $\endgroup$
    – Moo
    Jan 25 '16 at 5:57
  • $\begingroup$ I wonder if the equation could be instead $y'=(y^2-1)e^{t}$. As Moo commented, there is probably a typo in the question. $\endgroup$ Jan 25 '16 at 6:18
  • $\begingroup$ If, as others have suggested,there is a typo, and we have,instead, $dy/dt=(y^2-)e^t$ then we have$ d(-e^{-t})/dt=e^{-t}dt/dy=1/(y^2-1),$ giving $-e^{-t}=K+\int (y^2-1)^{-1}dy.$ $\endgroup$ Jan 25 '16 at 8:02


Let $u=ty$ ,

Then $y=\dfrac{u}{t}$


$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\left(\dfrac{u^2}{t^2}-1\right)e^u$ with $u(t=1)=0$

$\dfrac{1}{t}\dfrac{du}{dt}=\dfrac{u^2e^u+u}{t^2}-e^u$ with $u(t=1)=0$

$(u^2e^u+u-e^ut^2)\dfrac{dt}{du}=t$ with $t(u=0)=1$

Let $v=t^2$ ,

Then $\dfrac{dv}{du}=2t\dfrac{dt}{du}$

$\therefore\dfrac{u^2e^u+u-e^ut^2}{2t}\dfrac{dv}{du}=t$ with $v(u=0)=1$

$(u^2+ue^{-u}-t^2)\dfrac{dv}{du}=2e^{-u}t^2$ with $v(u=0)=1$

$(u^2+ue^{-u}-v)\dfrac{dv}{du}=2e^{-u}v$ with $v(u=0)=1$

Let $w=u^2+ue^{-u}-v$ ,

Then $v=u^2+ue^{-u}-w$


$\therefore w\left(2u+(1-u)e^{-u}-\dfrac{dw}{du}\right)=2e^{-u}(u^2+ue^{-u}-w)$ with $w(u=0)=1$

$(2u+(1-u)e^{-u})w-w\dfrac{dw}{du}=2ue^{-u}(u+e^{-u})-2e^{-u}w$ with $w(u=0)=1$

$w\dfrac{dw}{du}=(2u+(3-u)e^{-u})w-2ue^{-u}(u+e^{-u})$ with $w(u=0)=1$

This belongs to an Abel equation of the second kind.

In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $w=\dfrac{1}{z}$ ,

Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$

$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2u+(3-u)e^{-u}}{z}-2ue^{-u}(u+e^{-u})$ with $z(u=0)=1$

$\dfrac{dz}{du}=2ue^{-u}(u+e^{-u})z^3+((u-3)e^{-u}-2u)z^2$ with $z(u=0)=1$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

  • $\begingroup$ Could you please expand on what is the goal of these manipulations? $\endgroup$
    – Artem
    Feb 11 '16 at 20:53

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.