Question

Find the solution of the given initial value problem and draw its graph. \(y^{\prime \prime}+4 y=2 \delta(t-\pi / 4) ; \quad y(0)=0, \quad y^{\prime}(0)=0\)

Short answer

Answer: The solution to the given IVP is a constant function, \(y(t) = 0\).

Step-by-step solution

Find the homogeneous solution

For the homogeneous solution, we have the equation \(y^{\prime \prime} + 4y = 0\). The characteristic equation is \(r^2 + 4 = 0\). Solving for \(r\), we get \(r = \pm 2i\). Therefore, the homogeneous solution is given by \(y_h(t) = A\cos(2t) + B\sin(2t)\).

Apply Laplace transform

Now, we will apply the Laplace transform on both sides of the given IVP. \(\mathcal{L}\{y^{\prime \prime} + 4y\} = \mathcal{L}\{2\delta(t - \frac{\pi}{4})\}\) Using properties of Laplace transform, \(s^2Y(s) - sy(0) - y^{\prime}(0) + 4Y(s) = 2e^{-\frac{\pi s}{4}}\) Since \(y(0) = 0\) and \(y^{\prime}(0) = 0\), we have \((s^2 + 4)Y(s) = 2e^{-\frac{\pi s}{4}}\).

Find the inverse Laplace transform

To find the particular solution, we need to find the inverse Laplace transform of \(Y(s) = \frac{2e^{-\frac{\pi s}{4}}}{s^2 + 4}\). Using convolution theorem, \(y_p(t) = y_h(t)*g(t)\) where g(t) is the inverse Laplace transform of \(2e^{-\frac{\pi s}{4}}\). \(g(t) = \mathcal{L}^{-1}\{2e^{-\frac{\pi s}{4}}\}\) Since \(g(t) = 2\delta(t - \frac{\pi}{4})\), the inverse Laplace transform of \(g(t)\) is equal to the forcing function. Now, we find the convolution integral, \(y_p(t) = (A\cos(2t) + B\sin(2t))*(2\delta(t - \frac{\pi}{4}))\) \(y_p(t) = \int_{0}^{t} (A\cos(2(\tau - \frac{\pi}{4})) + B\sin(2(\tau - \frac{\pi}{4})))2\delta(\tau - \frac{\pi}{4}) d\tau\) Using the property of delta function, we can evaluate the integral as \(y_p(t) = 2(A\cos(\frac{\pi}{2}) + B\sin(\frac{\pi}{2}))\) \(y_p(t) = 2B\)

Apply initial conditions

To find the values of A and B, we will apply the given initial conditions. \(y(0) = 0 = A\cos(0) + B\sin(0)\) => \(A = 0\) \(y^{\prime}(0) = 0 = -2A\sin(0) + 2B\cos(0)\) => \(B = 0\) Since A and B are both 0, \(y_p(t) = 0\).

Combine the homogeneous and particular solutions

The general solution for the given IVP is the sum of homogeneous and particular solutions. \(y(t) = y_h(t) + y_p(t)\) \(y(t) = A\cos(2t) + B\sin(2t) + 0\) As A and B are both 0, the solution becomes \(y(t) = 0\)

Draw the graph of the solution

The graph of the solution is a horizontal line through the origin with the equation \(y(t) = 0\). Since it's a constant, the graph remains the same at all points in time.