Solving a Linear IVP

Question

I need help solving this linear Initial value problem: $$y'=-L(y(t)-\phi(t))+\phi'(t) \\ y(0)=y_0$$ where $\phi(t)=\cos(30t)$.

Answer 1

Rewriting the differential equation as $$y' + Ly = L\cos(30t) - 30\sin(30t), \qquad \qquad (1)$$ we first solve the homogeneous part of the differential equation $$y' + Ly = 0$$ and so have a complementary function of $$Y_{CF} = Ae^{-LT}.$$

We look for a particular integral that satisfies the differential equation of the form $y_{PI} = B\sin(30t)+C\cos(30t)$. We have $y'_{PI} = 30B\cos(30t) - 30C\sin(30t)$ and plugging into $(1)$

$$30B\cos(30t) - 30C\sin(30t) + LB\sin(30t)+LC\cos(30t) = L\cos(30t) - 30\sin(30t).$$

Comparing coefficients, $$30B + LC = L$$ $$LB - 30C = -30$$ $$\implies C = 1, \qquad B = 0.$$

So the general solution is the superposition of the complimentary function and the particular integral $$y_{GS} = Ae^{-Lt} + \cos(30t).$$

Finally we apply our initial condition to find the constant $A$.

$$y_{GS}(0) = A + 1 = y_0 \implies A = y_0 - 1.$$

This gives us the solution

$$y(t) = (y_0 - 1)e^{-Lt} + cos(30t),$$

as required.

Answer 2

Alternatively, again rewriting your differential equation as $$y' + Ly = L\cos(30t) - 30\sin(30t)$$

taking the Laplace transform and expanding into a partial fraction decomposition gives

$$Y(s) = \dfrac{y_0s^2 + Ls + 900y_0 - 900}{(s^2 + 900)(s+L)} = \dfrac{A}{s+L} + \dfrac{Bs+C}{s^2+900}.$$

$$\implies y_0s^2 + Ls + 900y_0 - 900 = A(s^2 + 900) + (Bs+C)(s+L)$$

Choosing $$s = -L: \quad y_0(L^2 + 900) - (L^2+900) = A(L^2+900), \quad \therefore \quad A = y_0 - 1$$ $$ s = 0: \qquad\qquad900(y_0 - 1) = 900(y_0 - 1) + LC \quad \qquad\qquad \therefore C = 0.$$

Finally comparing the $s^2$ coefficients we must have that $y_0 = y_0 - 1 + B \implies B = 1.$

So our decomposition is

$$Y(s) = \dfrac{y_0-1}{s+L} + \dfrac{s}{s^2+30^2};$$

and clearly from our tables

$$\mathcal{L}^{-1}(Y(s)) = y(t) = (y_0 - 1)e^{-Lt} + \cos(30t),$$

which tallies with the previous answer.