Determine the form of a particular solution for the given equation. Do not solve the equation.

y″−24y′+144y=t2e12t+e12t+4sin(12t)

Question

Determine the form of a particular solution for the given equation. Do not solve the equation.

y″−24y′+144y=t2e12t+e12t+4sin(12t)

BlackTom AI · Accepted Answer

The given differential equation is y'' - 24 y' + 144 y = t^2 e^{12t} + e^{12t} + 4 sin(12t).

First, identify the homogeneous solution structure. The characteristic equation is r^2 - 24 r + 144 = (r - 12)^2, so r = 12 is a repeated root of multiplicity 2. This means the complementary (homogeneous) solution includes terms like e^{12t} and t e^{12t}, i.e., y_h(t) = (C1 + C2 t) e^{12t}.

Next, determine the form of a particular solution for each forcing term on the right-hand side, keeping in mind resonance with the homogeneous solution.
- For the term t^2 e^{12t}: This is e^{rt} times a polynomial P(t) with r = 12. Since r = 12 is a root of the characteristic polynomial with multiplicity m = 2, we must multiply the usual ansatz by t^m to avoid duplication. The usual form for P(t) of degree 2 would be e^{12t}(A t^2 + B t + C), but due to resonance with multiplicity 2, we multiply by t^2, giving e^{12t} t^2 (A t^2 + B t + C) = e^{12t}(A t^4 + B t^3 + C t^2).
- For the term e^{12t} (the constant polynomial P(t) of degree 0): By the same reasoning, this becomes e^{12t} t^2 times a constant, which is encompassed in the A t^4 + B t^3 + C t^2 form above. So we can treat both t^2 e^{12t} and e^{12t} together under a single e^{12t} times a polynomial starting at t^2.
- For the term 4 sin(12t): This is a sinusoidal forcing with frequency 12. Since the homogeneous solution only involves e^{12t} (not sin/cos with imaginary roots), there is no resonance with sin(12t). The standard trial for a sin forcing of frequency 12 is a linear combination of cos(12t) and sin(12t): D cos(12t) + E sin(12t).

Combining these observations, a suitable form for a particular solution is:
- For the exponential-polynomial part: e^{12t}(a t^4 + b t^3 + c t^2)
- For the sinusoidal part: d cos(12t) + e sin(12t)

Therefore, a complete form to test as a particular solution (without solving the equation) is:
yp(t) = e^{12t}(a t^4 + b t^3 + c t^2) + d cos(12t) + e sin(12t).

If one prefers to separate contributions, one could also write:
yp(t) = e^{12t}(A t^4 + B t^3 + C t^2) + D cos(12t) + E sin(12t).

Note: Because the instruction says not to solve the equation, we stop at identifying the appropriate form for the particular solution and do not determine the coefficients A, B, C, D, E.

MAP2302 Lecture Quiz 15

Determine the form of a particular solution for the given equation. Do not solve the equation. y″−24y′+144y=t2e12t+e12t+4sin(12t)

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