I am a little confused how to solve these problems. They are under miscellaneous problems of chapter 2.9 in our textbook, but the second order differential equations don't start until chapter 3. The only instructions on how to solve these problems are provided below. I have to use some kind of substitution?
Equations with the Dependent Variable Missing. For a second order differential equation of the form y''=f(t,y'), the substitution v=y', v'=y'' leads to a first order equation of the form v'=f(t,v). If this equation can be solved for v, then y can be obtained by integrating dy/dt=v. Note that one arbitrary constant is obtained in solving the first order equation for v, and a second is introduced in the integration for y.
- $t^2y''+2ty'-1=0$, $t>0$ (Solution: $y = c_1t^{-1}+c_2+ln(t)$)
- $ty''+y'=1$, $t>0$ (Solution: $y = c_1ln(t)+c_2+t$)
- $t^2y''=(y')^2$ , $t>0$ (Solution: $c_1^2y=c_1t-ln|1+c_1t|+c_2$)
Update: Included solutions if that helps.
Best Answer
We can transform a second order differential equation of the form $y''=f(t,y')$, using the substitution $v=y', v'=y''$, to a first order equation of the form $v'=f(t,v)$.
Lets do $(3)$ as an example, so we have:
$t^2y''=(y')^2$ , $t>0$
Let: $v=y', v'=y''$, yielding (notice that we have a form $v' = f(t, v)$):
$$t^2 v' = (v)^2$$
We can easily separate this as:
$$\tag 1 \frac{dv}{v^2} = \frac{dt}{t^2}$$
Update
From $(1)$, we get:
$$\frac{1}{v} = \frac{1}{t} + c_1$$
This can be written as: $\displaystyle \frac{1}{v} = \frac{1 + c_1t}{t}$, and then we solve for $v$, yielding:
$$v = \frac{t}{1+c_1t}$$
However, from our earlier substitution, we have $v = y'$, so:
$$y' = \frac{t}{1+c_1t}$$
Can you solve for $y$ now?
Update 2
Note: when you solve for $y'$, you'll get another constant $c_2$ from that, to go along with the $c_1$. You would need to be given some ICs to figure those out - so they are left in this form.
Regards