Determine the form of a particular solution to

y″−10y′+26y=e2t+tsin5t−cos5t

Question

Determine the form of a particular solution to

y″−10y′+26y=e2t+tsin5t−cos5t

BlackTom AI · Accepted Answer

We start by clearly restating the problem and the given choices to ensure we’re evaluating the right forms.
Question: Determine the form of a particular solution to y'' − 10y' + 26y = e^{2t} + t sin(5t) − cos(5t).
Answer options:
1) yp = A cos(5t) + B t sin(5t) + C e^{2t}
2) yp = (A1 t + A0) cos(5t) + (B1 t + B0) sin(5t) + C e^{2t}
3) yp = (A1 t + A0) cos(5t) + (B1 t + B0) sin(5t) + C t e^{2t}
4) yp = t(A1 t + A0) cos(5t) + t(B1 t + B0) sin(5t) + C e^{2t}

Now, analyze each option in light of standard methods for linear ODEs with constant coefficients.
Option 1: yp = A cos(5t) + B t sin(5t) + C e^{2t}.
- This form does not include a term with t cos(5t), which is typically needed when the forcing contains sin(5t) terms that require a combination of sine and cosine components with polynomial in t. The absence of the t cos(5t) term suggests an incomplete ansatz for the t sin(5t) forcing, leading to an inadequate capture of the particular solution. Moreover, the e^{2t} forcing would usually yield a simple Ce^{2t} term, which is present, but the cos(5t) part is missing the necessary t-dependence to handle the t sin(5t) input effectively. Overall, this option is unlikely to satisfy all forcing components.
Option 2: yp = (A1 t + A0) cos(5t) + (B1 t + B0) sin(5t) + C e^{2t}.
- This is the standard form for a forcing that includes a cos(5t) and sin(5t) with linear t terms, i.e., a polynomial in t multiplying cosine and sine at frequency 5. Since the forcing contains t sin(5t) and −cos(5t), including linear terms in t for both cos(5t) and sin(5t) provides enough flexibility to satisfy those components when substituted into the differential equation. The Ce^{2t} term correctly accounts for the e^{2t} forcing, as 2 is not a root of the characteristic equation, so a simple exponential term is appropriate.
Option 3: yp = (A1 t + A0) cos(5t) + (B1 t + B0) sin(5t) + C t e^{2t}.
- The e^{2t} forcing would typically yield a Ce^{2t} term, not a t e^{2t}, because r = 2 is not a root of the characteristic equation (the roots are 5 ± i). Introducing a t e^{2t} factor unnecessarily increases the degree of the polynomial multiplying the exponential and is not required by the standard method unless resonance occurs. Therefore, this form overcompensates for the e^{2t} part and is not appropriate.
Option 4: yp = t(A1 t + A0) cos(5t) + t(B1 t + B0) sin(5t) + C e^{2t}.
- This option multiplies the cos and sin terms by t^2 in places, suggesting a quadratic in t factor. Since the forcing includes t sin(5t) and cos(5t) (not t^2 sin(5t) or t^2 cos(5t)), a quadratic-in-t multiplier is overly aggressive and not warranted by the standard method. It would generally introduce unnecessary higher-order terms and complicate the solution without necessity.

Taken together, option 2 provides the appropriate structure: a linear polynomial in t multiplying both cos(5t) and sin(5t) to accommodate the t sin(5t) and −cos(5t) components, plus a simple Ce^{2t} term for the e^{2t} forcing (since 2 is not a root of the characteristic equation). The other options either miss needed t-dependence, overcount t-powers, or misallocate the exponential part.

MAP2302 L15 section 4.5

Determine the form of a particular solution to y″−10y′+26y=e2t+tsin5t−cos5t

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