Why is the constant assumed to be zero when solving second order differential equation? Using the reduction of order method to solve a differential equation, let $y_1(t) = t^{-1}$ and $y_2(t) = v(t) \times t^{-1}$ 
I got $2tv''(t) - 3v'(t) =0$
Let $w(t) = v'(t)$ and I will get
$w(t) = ct^{\frac{3}{2}}$
And I can integrate $w(t)$ to find $v(t)$,
$$v(t) = \int w(t)dt = Ct^{\frac{5}{2}} + k$$
So, here's my problem, my maths teacher told me that the constant in $v(t)$ should be neglected, I need to assume $C = \frac{2}{5}$ and $k = 0$. But I don't understand why should I do this.
 A: The general solution is $c_1 y_1 + c_2 y_2$ for any constants $c_1,c_2$, where the $y_1,y_2$ are any two linearly independent solutions to the ODE. When you are applying reduction of order to get $y_2$ when you already know $y_1$, the required linear independence is achieved if $v$ isn't constant.
Since you're just looking for any solution $y_2$ linearly independent of $y_1$, you have the freedom to select the constants of integration arising in the reduction of order ODE (the $2tv''-3v'=0$ in your problem) to be essentially whatever you want. The only limitation is that you can't pick $C=0$, because then $v$ would be constant and so $y_1$ and $y_2$ wouldn't be linearly independent.
A: Ian give an answer for your question. Then, the general solution should be $y(t)=c_1t^{-2}+c_2t$ for arbitrary constant $c_1$ and $c_2$. Another way similar for the substitution is noticing that we have $t^2y''+2ty'-2y=0$ is an Euler-Cauchy equation, so we try $y:=t^{\lambda}$ then we can rewrite the ode as $(\lambda^2+\lambda-2)t^{\lambda}=0$. When $t\not=0$ we have the quadratic equation $\lambda^2+\lambda-2=0$ solving it we get  $\lambda=-2$ or $\lambda =1$. Therefore general solution is given by $y(t)=c_1t^{-2}+c_2t$ for arbitrary constants $c_1$ and $c_2$.
A: $$2tv''(t) - 3v'(t) =0$$
$$\dfrac {v''(t) }{ v'(t)}=\dfrac 3{2t}$$
$$ \ln v'=\dfrac 32 \ln t+C$$
$$v'=ct^{3/2}$$
$$v=C_1t^{3/2}+C_2$$
You can assume also that $C_2 \ne 0$ then the general solution is:
$$y= A y_1+By_2$$
$$y=A\dfrac 1t+B(v(t) y_1(t))$$
$$y=A\dfrac 1t+\dfrac B t(C_1t^{3/2}+C_2)$$
$$y=(A+BC_2)\dfrac 1t+\dfrac B tC_1t^{3/2}$$
$$y=(A+BC_2)\dfrac 1t+ B C_1t^{1/2}$$
More simply:
$$y=C\dfrac 1t+  C_0t^{1/2}$$
The constant $C_2$ is absorbed so setting it to zero is more simple.
