1. ## Undetermined Coefficients

I have a problem solving the undetermined coefficient problems. I know how to get the particular solution. But the algebra always messes me up.

The problem is:
y'' + 9y = t^2*e^3t + 6
The particular solution y particular = (At^2 + Bt + C)e^3t + D.

The next one is:
2y'' + 3y' + y = t^2 + 3sint
The particular solution y particular = (At^2 + Bt + C) + Asint + Bcost

This problem is an initial value problem:
y'' + 4y = 3sint2t, y(0) = 2, y'(0) = -1
The particular solution y particular was the same as the homogeneous solution so I added a factor t to the particular solution. So my y particular was: t(Asin2t + Bcos2t).

Tell me if these particular solutions are correct. My main problem was to solve for the constants like A, B, and even C. Can you show the step of how I can solve this. Even though I collected the terms but I get a strange equation that equals to the g(t).

2. Here's the second one
Originally Posted by fudawala
The next one is:
2y'' + 3y' + y = t^2 + 3sint
The particular solution y particular = (At^2 + Bt + C) + Asint + Bcost
$2y'' + 3y' + y = t^2 + 3 \sin t$

For the homogeneous solution, try $y = e^{ \lambda t}$, thus we get the characteristic equation:

$2 \lambda ^2 + 3 \lambda + 1 = 0$

$\Rightarrow \lambda = - \frac {1}{2} \mbox { or } \lambda = -1$

So, $y_h = Ae^{-t} + Be^{-t/2}$

For a particular solution, try $y_p = At^2 + Bt + C + D \sin t + E \cos t$ ........I hope you notice your mistake here. You reused A and B for the sine and cosine functions, you can't do that, they should be different variables!

$\Rightarrow y_p ' = 2At + B + D \cos t - E \sin t$

$\Rightarrow y_p'' = 2A - D \sin t - E \cos t$

Plug these into the original equation, we get:

$4A - 2D \sin t - 2E \cos t + 6At + 3B + 3D \cos t - 3E \sin t + At^2$ $+ Bt + C + D \sin t + E \cos t = t^2 + 3 \sin t$

Collect like terms:

$\Rightarrow (4A + 3B + C) + (6A + B)t + At^2 + (-3E - D) \sin t + (3D - E) \cos t = t^2 + 3 \sin t$

Equating coefficients we get:

$\boxed { A = 1}$

$\Rightarrow 3B + C = -4$

$\boxed { B = - 6} \implies \boxed { C = 14 }$

$-D - 3E = 3$

$3D - E = 0 \implies E = 3D$

$\Rightarrow -D - 9D = 3 \implies \boxed { D = - \frac {3}{10} } \implies \boxed { E = - \frac {9}{10}}$

Thus the general solution is:

$y_g = y_h + y_p$

$\Rightarrow y_g = Ae^{-t} + Be^{-t/2} + t^2 - 6t + 14 - \frac {3}{10} \sin t - \frac {9}{10} \cos t$

3. Here's the first

Originally Posted by fudawala

The problem is:
y'' + 9y = t^2*e^3t + 6
The particular solution y particular = (At^2 + Bt + C)e^3t + D.
Good job getting the particular solution for that, now let's see how it all works out.

$y'' + 9y = t^2 e^{3t} + 6$

For the homogeneous solution, try $y = e^{ \lambda t}$, thus we get the characteristic equation:

$\lambda ^2 + 9 = 0$

$\Rightarrow \lambda = \pm 3~i$

$\Rightarrow y_h = A \cos 3t + B \sin 3t$

For the particular solution, try $y_p = \left( At^2 + Bt + C \right)e^{3t} + D$

$\Rightarrow y_p' = (2At + B) e^{3t} + 3 \left( At^2 + Bt + C \right) e^{3t}$

$\Rightarrow y_p'' = 2Ae^{3t} + 3(2At + B)e^{3t} + 3(2At + B)e^{3t} + 9 \left( At^2 + Bt + C \right)e^{3t}$

Plug these into the original equation (you may simplify them first if you wish, but I'll jump right into it, I will also expand the brackets at the same time, remember, we must multiply $y_p$ by 9), we get:

$2Ae^{3t} + 6Ate^{3t} + 3Be^{3t} + 6Ate^{3t} + 3Be^{3t} + 9At^2 e^{3t} + 9Bte^{3t} + 9C e^{3t} +$ $9At^2 e^{3t} + 9Bte^{3t} + 9Ce^{3t} + 9D = t^2 e^{3t} + 6$

Collect like terms:

$\Rightarrow (2A + 6B + 18C)e^{3t} + (12A + 18B)te^{3t} + 18At^2 e^{3t} + 9D = t^2 e^{3t} + 6$

Equating coefficients, we get:

$2A + 6B + 18C = 0$

$12A + 18B = 0$

$18A = 1 \implies \boxed { A = \frac {1}{18} }$ $\implies \boxed { B = - \frac {1}{27} }$

plug A and B into the first equation, we get $\boxed { C = \frac {1}{162} }$

$9D = 6 \implies \boxed { D = \frac {2}{3} }$

So our general solution is:

$y_g = y_h + y_p$

$\Rightarrow y_g = A \cos 3t + B \sin 3t + \left( \frac {1}{18} t^2 - \frac {1}{27} t + \frac {1}{162} \right)e^{3t} + \frac {2}{3}$

Or

$y_g = A \cos 3t + B \sin 3t + \frac {1}{162} \left( 9t^2 - 6t + 1 \right)e^{3t} + \frac {2}{3}$

4. Originally Posted by fudawala
This problem is an initial value problem:
y'' + 4y = 3sint2t, y(0) = 2, y'(0) = -1
The particular solution y particular was the same as the homogeneous solution so I added a factor t to the particular solution. So my y particular was: t(Asin2t + Bcos2t).
Okay, let's see how far we get with this one.

$y'' + 4y = 3 \sin 2t$ $y(0) = 2 \mbox { , } y'(0) = -1$

For the homogeneous solution, try $y = e^{ \lambda t}$, thus we get the characteristic equation:

$\lambda ^2 + 4 = 0$

$\Rightarrow \lambda = \pm 2i$

$\Rightarrow y_h = A \cos 2t + B \sin 2t$

For the particular solution we guess $y_p = At \sin 2t + Bt \cos 2t$

$\Rightarrow y_p' = A \sin 2t + 2At \cos 2t + B \cos 2t = 2Bt \sin 2t$

$\Rightarrow y_p'' = 2A \cos 2t + 2A \cos 2t - 4At \sin 2t - 2B \sin 2t - 2B \sin 2t - 4Bt \cos 2t$

Plugging these into the original equation, we get:

$2A \cos 2t + 2A \cos 2t - 4At \sin 2t - 2B \sin 2t - 2B \sin 2t -$ $4Bt \cos 2t + 4At \sin 2t + 4Bt \cos 2t = 3 \sin 2t$

This simplifies to:

$4A \cos 2t - 4B \sin 2t = 3 \sin 2t$

Equating the coefficients we get:

$4A = 0 \implies \boxed { A = 0 }$

$-4B = 3 \implies \boxed { B = - \frac {3}{4} }$

Thus, our general solution is of the form:

$y_g = y_h + y_p$

$\Rightarrow y_g = y(t) = A \cos 2t + B \sin 2t - \frac {3}{4} t \cos 2t$

$\Rightarrow y'(t) = -2A \sin 2t + 2B \cos 2t - \frac {3}{4} \cos 2t + \frac {3}{2} \sin 2t$

we are given the initial data that $y(0) = 2 \mbox { and } y'(0) = -1$

$\Rightarrow y(0) = A = 2 \implies \boxed { A = 2}$

$\Rightarrow y'(0) = 2B - \frac {3}{4} = -1 \implies \boxed { B = - \frac {1}{8}}$

And so, finally, our general solution is:

$y(t) = 2 \cos 2t - \frac {1}{8} \sin 2t - \frac {3}{4} t \cos 2t$

Now that wasn't so bad, was it?

5. ## Method of Undetermined Coefficients

Hey Jhevon,

The question that you solved:
2y'' + 3y' + y = t^2 + 3 \sin t

How did you equate the equation 3B + C = -4. I don't understand where did you get that from.

6. Originally Posted by fudawala
Hey Jhevon,

The question that you solved:
2y'' + 3y' + y = t^2 + 3 \sin t

How did you equate the equation 3B + C = -4. I don't understand where did you get that from.
Notice that after we collected the terms and wanted to equate coefficients, we had $4A + 3B + C$ being equated to the constant term on the right side of the equation, however, there was no constant term on the right. thus we have:

$4A + 3B + C = 0$

i found that $A = 1$ from another term right of the bat, so we now have

$4 + 3B + C = 0$

subtracting 4 from both sides we obtain

$3B + C = -4$

which is the equation in question

sorry i didn't show all the steps, these problems are too much typing to begin with, i got lazy