How Do You Solve a Nonhomogeneous Second-Order Differential Equation?

[Stu165]

Can anyone give me a hand with this, cause I'm stumped and can't remember exactly how to go about solving this.

here's the eqn

m[d^2x/dt^2 + wsubo^2 x] = F cos wt

I'm supposed to show that x(t) = xsubo cos wt

w is the incident freq
wsubo is the resonant freq
m is mass

I'm stuck with the part of getting out ysubp, at least that's what I think I should do.

 

[dextercioby]

Can u solve the homogenous equation...?If so,then u can use the method of Lagrange to find a particular solution of the nonhomogenous ODE.

Daniel.

 

[thepatientmental]

Sure, I can give you a hand with this. The equation you have is a second-order nonhomogeneous differential equation, meaning that it involves a second derivative of the dependent variable (x) and has a non-zero right-hand side (F cos wt). To solve this type of equation, we use the method of undetermined coefficients.

First, we need to find the complementary solution, which is the solution to the homogeneous equation (without the right-hand side). In this case, the homogeneous equation is m(d^2x/dt^2 + wsubo^2 x) = 0. The solution to this equation is given by x(t) = A cos(wsubo t) + B sin(wsubo t), where A and B are arbitrary constants.

Next, we need to find the particular solution, which is a specific solution that satisfies the nonhomogeneous equation. In this case, we can assume that the particular solution has the form x(t) = Asin(wt) + Bcos(wt), where A and B are again arbitrary constants. This is because the right-hand side of the equation is a cosine function, and our complementary solution already contains a cosine function.

Plugging this particular solution into the original equation, we get:

m(d^2/dt^2 + wsubo^2)(Asin(wt) + Bcos(wt)) = F cos wt

Expanding and simplifying, we get:

-mAw^2sin(wt) + mBw^2cos(wt) + mwsubo^2Asin(wt) + mwsubo^2Bcos(wt) = F cos wt

Equating coefficients of like terms, we get:

-mAw^2 + mwsubo^2A = 0 and mBw^2 + mwsubo^2B = F

Solving for A and B, we get:

A = F/(mwsubo^2 - mAw^2) and B = 0

Therefore, the particular solution is x(t) = F/(mwsubo^2 - mAw^2) sin(wt).

Finally, the general solution to the nonhomogeneous equation is given by the sum of the complementary and particular solutions:

x(t) = A cos(wsubo t) + B sin(wsubo t) + F/(mwsubo^2 - mAw^2) sin(wt)