Finding solution of $y' = \frac{(y^2 - 4yt + 6t^2)}{t^2}$ with initial condition $y(2) = 4$

Question

Concerning the ordinary differential equation $$y' = \frac{(y^2 - 4yt + 6t^2)}{t^2}$$ with initial condition $y(2) = 4$.

My text book gives the solution as $y = 2t$, and so does MATLAB.

However, I only see the following :
\begin{align} v &= \frac{y}{t}\\ y' &= v^2 - 4v + 6\\ tv' + v &= v^2 - 4v + 6\\ tv'&= v^2 - 5v + 6\\ \end{align}

\begin{align} \frac{dv}{(v-3)(v-2)} &= \frac{dt}{t}\\ \frac{dv}{(v-3)} - \frac{dv}{(v-2)} &= \frac{dt}{t}\\ \ln(|v-3|) - \ln(|v-2|) &= \ln(|t|) + c\\ \frac{v-3}{v-2} = te^{c}\\ \end{align} which then reduces to
$$v = \frac{3 - 2te^{c}}{1-te^{c}}$$ and $$y = \frac{t(3 - 2te^{c})}{1-te^{c}}$$

There must be at least some logic to this, because several online calculators (for example WolframAlpha) give the same result. But nevertheless it seems wrong, as the initial condition implies
$4 - 8e^{c} = 6 - 8e^{c}$?

Can someone please explain to me how to arrive at $y=2t$, and where I would have gone wrong?

Answer 1

We can rewrite the differential equation in the following way

$$y' - 2 = \frac{(y-2t)^2}{t^2}$$

where we can now use the substitution $v = y-2t$

$$v' = \frac{v^2}{t^2} \implies v = \frac{t}{1+Ct}$$

Then plug in the initial condition $v(2) = y(2)-2\cdot 2 = 0$. However, notice that the "general solution" has a singularity there. The resolution is that we assumed $v$ was nonzero when we separated variables, which we can't do. Fortunately for us, $v(t) = 0$ is a solution to the original ODE. This means that

$$y(t) = v(t) + 2t = 2t$$

Answer 2

You've shown

$$ tv' = v^2 - 5v + 6 = (v-2)(v-3).$$

and since $v = y/t$, the initial condition gives $v(2) =2$. Thus $$\frac{dv}{(v-3)(v-2)} = \frac{dt}{t}$$

cannot be done since $v-2$ is at the bottom.

This is exactly the case where sometimes separation of variable does not work: indeed for a simple problem like

$$ y' = y, \ \ \ y(0) = 0,$$ you cannot use separation of variable to find the solution (which is $y(t) = 0$).

To "find" the solution $y = 2t$, you simply observe that $v(t) = 2$ is a solution, and then from the uniqueness theorem to ODE, it is THE solution.

Answer 3

$$y' = \frac{(y^2 - 4yt + 6t^2)}{t^2}$$ $$y't^2=(y-2t)^2+2t^2$$ $$t^2(y'-2)=(y-2t)^2$$ $$t^2(y-2t)'=(y-2t)^2$$ Substitute $z=y-2t$ $$t^2z'=z^2 \implies z'=\left ( \dfrac zt \right)^2=f(z,t)$$ With intial condition: $$z(2)=0$$ Since $z(t)=0$ is a solution and by the uniqueness theorem ($f$ and $\partial_z f$ are continuous on $t \in (0,\infty]$), the only solution is therefore $z(t)=0 \implies y(t)=2t$.