About differentiable dependence in a Cauchy problem

Question

Let $\epsilon >0$, consider the Cauchy problem: $$\epsilon x' = x^2 + (1-\epsilon)t \quad , x(0)=1$$ If $x(t;\epsilon)$ denotes the solution (defined on the maximum interval) of the problem, I'm asked to verify that: $$\frac{\partial x}{\partial \epsilon} (t,1) = \frac{t}{(1-t)^2} \left ( \frac{t^2}{3} - \frac{t}{2} -1 \right )$$ I'm not able to get the previous expression and I'm getting kinda crazy xd. I know for a fact that the function $u = \partial_\epsilon (\cdot,1)$ solves the linear problem: $$u' = f_x (t,x(t;1),1) u + f_\epsilon (t,x(t;1),1) \quad , u(0)=0$$ In this case $f(t,x,\epsilon) = \frac{x^2}{\epsilon} + \frac{1-\epsilon}{\epsilon} t$ so: $$f_x (t,x,\epsilon) = \frac{2x}{\epsilon} \qquad f_\epsilon (t,x,\epsilon) = - \frac{x^2}{\epsilon^2} - \frac{t}{\epsilon^2}$$ , and the Cauchy problem I should work on is: $$u' = \frac{2}{1-t}u -t - \frac{1}{(1-t)^2} \quad , u(0)=0$$ , since $x(t;1) = (1-t)^{-1}$. The solution to this IVP is: $$u(t) = - \frac{t (3t^3 - 8t^2 +6t+12)}{12 (1-t)^2}$$ (by Wolfram-Alpha), and it doesn't match with the solution given. Is there any mistake in my work?. ANY suggestion will be appreciated :)

Answer 1

Taking directly the $\epsilon$ derivative of the original equation gives at $\epsilon=1$ $$ x'+u'=2xu-t,~~\text{ while }~~ x'=x^2 $$ which reproduces the equation that you found, $$ u'=2xu-x^2-t. $$ Applying the usual solution methods $$ \implies ((1-t)^2u)'=-1-t(1-t)^2=-1-(1-t)^2+(1-t)^3 \\ (1-t)^2u=-t+\frac13(1-t)^3-\frac14(1-t)^4-\frac1{12}=-\frac{t^4}4+\frac{2t^3}3-\frac{t^2}2-t $$ gives again what you found.

So your reference solution is wrong or is the solution to a different task.