Need help with solving an 2nd Order Linear ODE

[Turion]

Homework Statement

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0$$

Homework Equations

The Attempt at a Solution

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0\\ \\ \\ x_{ c }={ c }_{ 1 }coswt+{ c }_{ 2 }sinwt\\ \\ \\ { x }_{ p }=Atsinwt+Btcoswt\\ { x }_{ p }'=wAtcoswt-wBtsinwt\\ { x }_{ p }''=-{ w }^{ 2 }Atsinwt-{ w }^{ 2 }Btcoswt\\ \\ \\ Sub\quad { x }_{ p },\quad { { x }_{ p } }',\quad and\quad { { x }_{ p } }''\quad into\quad the\quad DE\\ \\ \\ 0={ F }_{ 0 }sinwt$$

I'm not sure where to proceed from here.

 

[pasmith]

Homework Statement

$$\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +{ w }^{ 2 }x={ F }_{ 0 }sinwt\quad \quad \quad \quad x(0)=0\quad \quad x'(0)=0$$

If you have
$$x'' + ax' + bx = f(t)$$
and the obvious particular solution candidate $g(t)$ is a solution of $x'' + ax' + bx = 0$, then try the candidate $tg(t)$.

 

[Turion]

If you have
$$x'' + ax' + bx = f(t)$$
and the obvious particular solution candidate $g(t)$ is a solution of $x'' + ax' + bx = 0$, then try the candidate $tg(t)$.

I don't follow. A particular solution applies to heterogeneous differential equations but you're using the term with a homogeneous differential equation?

 

[STEMucator]

I don't follow. A particular solution applies to heterogeneous differential equations but you're using the term with a homogeneous differential equation?

If a term appears in your complementary solution and then in your particular solution, your particular solution needs to be changed a bit. If the term is already in the complementary solution, how can it possibly be part of the particular solution?

That's why a useful trick is to multiply the guessed particular solution by the independent variable ( usually $t$ ), which will yield a completely different particular solution. If that does not get rid of all the duplicate terms between your complementary solution and your particular guess, multiply the particular guess by the independent variable a second time.

If you're working with a second order ODE, you will never need to multiply more than twice ( i.e the highest power term introduced through multiplying would be $t^2$ ). If you're working with an nth order ODE, you never need to multiply more than $n$ times.

For an easy example, try solving $y'' - 3y' - 4y = e^{-t}$. Your complementary solution won't be too bad to solve, but try guessing the particular solution as you did before without multiplying through by the independent variable. You'll notice no combination of coefficients will satisfy your equation and your particular solution guess will be incorrect.

 

[Turion]

If a term appears in your complementary solution and then in your particular solution, your particular solution needs to be changed a bit. If the term is already in the complementary solution, how can it possibly be part of the particular solution?

That's why a useful trick is to multiply the guessed particular solution by the independent variable ( usually $t$ ), which will yield a completely different particular solution. If that does not get rid of all the duplicate terms between your complementary solution and your particular guess, multiply the particular guess by the independent variable a second time.

If you're working with a second order ODE, you will never need to multiply more than twice ( i.e the highest power term introduced through multiplying would be $t^2$ ). If you're working with an nth order ODE, you never need to multiply more than $n$ times.

I did do that though:

$${ x }_{ p }=Atsinwt+Btcoswt$$

or am I missing something?

 

[pasmith]

I did do that though:



or am I missing something?

I seem to have missed the t's. You however missed them when differentiating:

$$(t \cos(wt))' = \cos(wt) - wt \sin(wt)$$

etc.

 

[Turion]

Thanks. Solved but was incredibly long.