Question

Consider the second order ordinary differential equation, y" +4y' +5y = 0. If y(0) = 0 and y'(0) = 1, then the value of у(π/2) is _________. (Round off to 3 decimal places).

Answer 1

Correct Answer: 0.041

To solve the differential equation \(y'' + 4y' + 5y = 0\) with initial conditions \(y(0) = 0\) and \(y'(0) = 1\), we first find the characteristic equation: \(r^2 + 4r + 5 = 0\). Solving by the quadratic formula, \(r = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a=1\), \(b=4\), \(c=5\), we get \(r = \frac{-4 \pm \sqrt{16-20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = -2 \pm i\). This gives us a solution form \(y(t) = e^{-2t}(C_1\cos t + C_2\sin t)\).

Applying initial conditions: \(y(0) = 0\) implies \(C_1 = 0\), as \(e^{0}(C_1\cos 0 + C_2\sin 0) = 0\). Then \(y'(t) = e^{-2t}(-2C_2\sin t + C_2\cos t)\), and \(y'(0) = 1\) gives us \(1 = C_2\). Thus, the particular solution is \(y(t) = e^{-2t}\sin t\).

Calculating \(y(\pi/2) = e^{-2(\pi/2)}\sin(\pi/2) = e^{-\pi}\). Calculating \(e^{-\pi} \approx 0.041\) to three decimal places, which lies in the specified range, 0.041,0.041.

Thus, \(y(\pi/2) \approx 0.041\).