1. ## Hard Dif. Equations

I have to solve the given Dif. EQs with the initial conditions that are given.

But, the hard part is I'm not allowed to use the same method twice. (That is, I can only use separation of variables once, variation of parameters once, etc).

1.) $(1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2$

2.) $y''' - y'' - 2y' = e^{-t} + 3e^{2t}, y(0) = 1, y'(0) = \frac{-13}{6}, y''(0) = \frac{-5}{3}$

3.) $\frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2$

4.) $y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0$

2. Originally Posted by fifthrapiers
1.) $(1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2$
This is a linear ODE.

Do you know how to do one of these?

3. Originally Posted by fifthrapiers
3.) $\frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2$
Factorise the denominator, then you'll be able to solve the separable ODE.

4. Originally Posted by Krizalid
Factorise the denominator, then you'll be able to solve the separable ODE.
Ahh, I definitely over looked this. Okay, does this look right:

$\displaystyle \frac{dy}{dt}=\frac{t^2}{y+t^3y}=\frac{t^2}{(t+1)( t^2-t+1)y}$

$\displaystyle y\cdot dy = \frac{t^2}{(t+1)(t^2-t+1)y}\cdot dt$

Take integral of both sides:

$\displaystyle \frac{y^2}{2} = \frac{\ln(|(t+1)(t^2-t+1)|)}{3} + C$

Use initial condition y(0) = -2

$\displaystyle \frac{(-2)^2}{2} = \frac{\ln(|(0+1)((0)^2-0+1)|)}{3} + C$

And so the particular solution, given the initial conditions is:

$\displaystyle \frac{y^2}{2} = \frac{\ln(|(t+1)(t^2-t+1)|)}{3} + 2$

5. Originally Posted by fifthrapiers
I have to solve the given Dif. EQs with the initial conditions that are given.

But, the hard part is I'm not allowed to use the same method twice. (That is, I can only use separation of variables once, variation of parameters once, etc).

1.) $(1 + t^2)\frac{dy}{dt} = 2ty + 2, y(0) = -2$

2.) $y''' - y'' - 2y' = e^{-t} + 3e^{2t}, y(0) = 1, y'(0) = \frac{-13}{6}, y''(0) = \frac{-5}{3}$

3.) $\frac{dy}{dt} = \frac{t^2}{y + t^3y}, y(0) = -2$

4.) $y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0$
I solved #2 using method of undetermined coefficients, so I don't need help on that anymore. Now I have #1 and #4 left (and if someone could check my #3). For #1, I think I have to use an integrating factor.....but not sure how to separate it into standard form first.. $y'(1 + t^2) - 2 = 2ty$ ...

That leaves me with #4... how can I solve this without using undetermined coefficients, separation of variables and integrating factor.. there's a method that is something involving constant coefficients, but not sure how to use that. I think I find y_c first and then y_p after... I can easily find y_c..its trying to find y_p that's hard.

6. Originally Posted by fifthrapiers
4.) $y'' + 4y' + 4y = t^{-2}e^{-2t}, y(1) = 0, y'(1) = 0$
No need to use variation of parameters use. Use the uniqueness theorem, notice that $y = 0$ for $t>0$ solves this equation trivial including the initial value problem. But the uniqueness theorem this is the only solution.

7. $\left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}
{{1 + t^2 }}y = \frac{2}
{{1 + t^2 }}$

So our I.F. will be $\exp \left( { - \int {\frac{{2t}}
{{1 + t^2 }}\,dt} } \right) = \left( {1 + t^2 } \right)^{ - 1}$

Multiply the equation by this I.F., then I assume you know you have to do.

8. Originally Posted by Krizalid
$\left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}
{{1 + t^2 }}y = \frac{2}
{{1 + t^2 }}$

So our I.F. will be $\exp \left( { - \int {\frac{{2t}}
{{1 + t^2 }}\,dt} } \right) = \left( {1 + t^2 } \right)^{ - 1}$

Multiply the equation by this I.F., then I assume you know you have to do.
Thanks a lot Kriz. I need some help finishing the integrating factor one:

$\left( {1 + t^2 } \right)y' - 2ty = 2 \iff y' - \frac{{2t}}{{1 + t^2 }}y = \frac{2}{{1 + t^2}}$

And so our integrating factor is:

$M(t) = e^{-\int{\frac{2t}{1 + t^2 }\,dt}} = e^{-\ln(x^2+1)} = \left( {1 + t^2} \right)^{-1}$

Multiply $M(t)$ by both sides of the equation.

$\left( {1 + t^2 } \right)y'\underbrace{\left( {1 + t^2} \right)^{-1}} -\, 2ty\underbrace{\left( {1 + t^2} \right)^{-1}} = 2\underbrace{\left({1 + t^2}\right)^{-1}}$

Now this is where things get fuzzy. Is it just:

$\left[\left( {1 + t^2 } \right)y\left( {1 + t^2} \right)^{-1}\right]' = 2\left({1 + t^2}\right)^{-1}$

And then take integrals of both sides (which the integral of the derivative on the left will just leave what that is, so I just have to integrate the right side)...or is this step wrong?

9. Can anyone confirm if this is right or help me in the right direction?