Show that separating the variables results in $v'(t) = ct^{1/2}$

Question

I have to solve the following exercise:

Suppose that we have the following differential equation: $$2tv'' - v' = 0$$ Show that separating the variables results in $v'(t) = ct^{1/2}$.

I understand separation of variables in the case of $\dfrac{dy}{dx} = f(y)f(x)$. We get $\dfrac{dy}{f(y)} = f(x)dx$. I have no idea however, how separation of variables solves the exercise. We have $2tv'' - v' = 0 \Rightarrow 2t\dfrac{d^2v}{dt^2} = \dfrac{dv}{dt}$. I have no clue how I should arrive at $v'(t) = ct^{1/2}$.

Question: How do I solve this exercise?

Answer 1

Our DE is

$$2tv'' = v'$$

Substitute $z = v'$

$$2tz' = z$$

Divide both sides by $2tz$

$$\frac{z'}{z} = \frac{1}{2t}, \quad t \neq 0, \quad z \neq 0$$

Integrate both sides

$$ln(z) + c_1 = \frac{1}{2}ln(t) + c_2$$

$$e^{ln(z) + c_1} = e^{ln(t^{\frac{1}{2}}) + c_2}$$

$$e^{c_1}z = e^{c_2}t^{\frac{1}{2}}$$

Substitute $v'$ back to get $v'(t) = e^{c_2 - c_1}t^{\frac{1}{2}}$

Answer 2

One thing to note about this problem is that v(t) never appears, neither in the problem statement, nor in what they ask you to prove. Therefore, the problem is simplified if you solve for v'(t) rather than v(t). So:

$$2t \frac{d(v')}{dt} = v'$$ $$\frac{d(v')}{v'} = \frac{t}{2dt}$$ $$ln(v') = \frac{ln(t)}{2}+c_1$$ $$ln(v') = ln(e^{c_1}t^{\frac{1}{2}})$$ $$v'=e^{c_1}t^{\frac{1}{2}}$$

If we take $c = c_1$, we get $v'=ct^{\frac{1}{2}}$

This can also be written as substitution, e.g. u=v', solve for u, then put v' back in for u.