The solution to the given differential equation can be expressed as y = Yivp + (h(t)*f(t)), where Yivp is the solution to the homogeneous equation, h(t) is the impulse response, and f(t) is the forcing function.
The solution to the differential equation y'' – 2y’ – 15y = -8t²e⁻³ᵗ with y(0) = 6, y’(0)=6 can be written in the form:
y = Yivp + (h(t)*f(t))
where:
Yivp is the solution to the associated homogeneous differential equation with the given initial values (ignore the forcing function, keep initial values).
h(t) is the impulse response (ignore the initial values and forcing function).
f(t) is the forcing function. (ignore the initial values and differential equation).
The first step is to find the solution to the homogeneous differential equation y'' – 2y’ – 15y = 0. This can be done using the characteristic equation:
r² – 2r – 15 = 0
The roots of this equation are r = -5 and r = 3. Therefore, the general solution to the homogeneous differential equation is:
Yivp = C₁e⁻⁵ᵗ + C₂e³ᵗ
we need to find the impulse response h(t). The impulse response is the solution to the differential equation y'' – 2y’ – 15y = δ(t), where δ(t) is the Dirac delta function. The Laplace transform of the impulse response is:
L{h(t)} = 1 / (s² – 2s – 15)
Taking the inverse Laplace transform, we find:
h(t) = (e⁻⁵ᵗ – e³ᵗ) / 4
Finally, we need to find the forcing function f(t). The forcing function is the right-hand side of the differential equation, which in this case is -8t²e⁻³ᵗ.