problem in a basic initial value problem $ \begin{cases} \ y'=t(y-1)^2\\ \ y(0)=0\\ \end {cases} $

Question

part one:$ \begin{cases} \ y'=t(y-1)^2\\ \ y(0)=0\\ \end {cases} $

$y(2)=?$

part two is similar but: $ \begin{cases} \ y'=t(y-1)^2\\ \ y(0)=1\\ \end {cases} $

also need to find $y(2)$

---

I solved it using the separable equation method

$\frac{dy}{dt}=t(y-1)^2$ $\iff$ $\int \frac{dy}{(y-1)^2}=\int tdt$ $\iff$ $-\frac{1}{y-1}=\frac{t^2}{2}+C$ now plugging in our initial value $\frac{2}{0-1}=0+C$ we get $C=-2$

so out function is $\frac{2}{y-1}=-t^2-2$ now to find $y(t=2)$ we have $\frac{2}{y-1}=-4-2$ $\iff$ $\frac{2}{-6}=y-1$ and finally $y=\frac{2}{3}$

for part two it is obviously the same integral so we have $-\frac{1}{y-1}=\frac{t^2}{2}+C$

but I cannot plug in $y(0)=1$ because then I get $\frac{1}{1-1}=\frac{1}{0}$

what am I missing?

Thanks for any help and tips!

Answer 1

You have done correctly for the first IVP.

The second IVP has no non constant solution.

---

Edit: Thanks to @conan for pointing out my fatal mistake.In the first place to solve the integration you have divided by $y-1$ which is valid only for $y(t) \neq 1$.

But $y(t) =1 $ solves the ode as well as satisfy the initial condition.

Hence $y(t) =1$ is the solution of the IVP.

But it doesn't have any non constant solution!