# Finding steady-state solution to 2nd-order DE

## Homework Statement

I want to get the particular solution for:

Lq''+Rq'+(1/c)q = Esin(xt)

## Homework Equations

Lq''+Rq'+(1/c)q = Esin(xt)

## The Attempt at a Solution

I got the particular solution and its derivatives:

q = A*Sin(xt) + B*cos(xt)
q' = x*A*cos(xt) - x*B*sin(xt)
q'' = -x^2*A*sin(xt) - x^2*B*cos(xt)

I plug these into the 2nd order equation giving me:

A[-x^2*L*sin(xt) + R*x*cos(xt) + (1/c)*Sin(xt)] + B[-x^2*L*cos(xt)-x*R*sin(xt)+(1/c)*cos(xt)] = Esin(xt)

I want to get the form of A and B in the attached image

Also, alpha = x in my typing.

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Orodruin
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Homework Helper
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Identifythe factors of sin(xt) on both sides of the wquation and then do the same for cos(xt). That gives you two equations for two unknowns.

Identifythe factors of sin(xt) on both sides of the wquation and then do the same for cos(xt). That gives you two equations for two unknowns.
I'm not sure if I understand. So I factor out sin(xt) on both sides and factoring cos(xt) giving me:

sin(xt)*[-L*x^2*A - R*x*B + (1/c)*A] + cos(xt)*[-L*x^2*B + R*x*A + (1/c)*B]

Orodruin
Staff Emeritus
Homework Helper
Gold Member
That's just one side of the equation. You need to do it on both sides and then identify the factors on either side.

That's just one side of the equation. You need to do it on both sides and then identify the factors on either side.
But the other side is just Esin(xt).

So I have:

(-L*x^2*B) + (R*x*A)+([1/c]*B) = 0
(-L*x^2*A) - (R*x*B) + ([1/c]*A) = E

Orodruin
Staff Emeritus