Solving the differential equation $y'=(y^2-1)e^{ty}$ 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?
Hint:

Let $u=ty$ ,

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

$\dfrac{dy}{dt}=\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}$

$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\left(\dfrac{u^2}{t^2}-1\right)e^u$ 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 $v=u^2+ue^{-u}-w$

$\dfrac{dv}{du}=2u+(1-u)e^{-u}-\dfrac{dw}{du}$

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

Let $u=ty$ ,

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

$\dfrac{dy}{dt}=\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}$

$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\left(\dfrac{u^2}{t^2}-1\right)e^u$ 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 $v=u^2+ue^{-u}-w$

$\dfrac{dv}{du}=2u+(1-u)e^{-u}-\dfrac{dw}{du}$

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