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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.

  1. $t^2y''+2ty'-1=0$, $t>0$ (Solution: $y = c_1t^{-1}+c_2+ln(t)$)
  2. $ty''+y'=1$, $t>0$ (Solution: $y = c_1ln(t)+c_2+t$)
  3. $t^2y''=(y')^2$ , $t>0$ (Solution: $c_1^2y=c_1t-ln|1+c_1t|+c_2$)

Update: Included solutions if that helps.

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See Update 2 for the last constant $c_2$. –  Amzoti Mar 4 '13 at 6:32

1 Answer 1

up vote 3 down vote accepted

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

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I integrated both sides and got $v^{-1}=t^{-1}+c$. That seems to be not even close to the solution from the textbook. –  Blue Pony Inc. Mar 4 '13 at 6:04
    
@BluePonyInc.: recall, you substituted, so you have to work all the way up the food chain to get back to the solution. This was the first step. Clear? –  Amzoti Mar 4 '13 at 6:05
    
I am still unsure. Could you please just finish problem #3 so I could see how it's done from start to finish and try to do the rest myself based off (3). –  Blue Pony Inc. Mar 4 '13 at 6:09
    
@BluePonyInc.: see update. –  Amzoti Mar 4 '13 at 6:19
1  
See update - just use a common denominator of $t$ on the right hand side of the equation, and then solve for $v$. Have fun with the other tools - now you have all the tools needed to do those! Regards –  Amzoti Mar 4 '13 at 6:28

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