Question

find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing\(t.\) $$ y^{\prime \prime}+y^{\prime}+1.25 y=0, \quad y(0)=3, \quad y^{\prime}(0)=1 $$

Short answer

The solution to the given second-order linear homogeneous differential equation with initial conditions is: $$ y(t) = e^{-\frac{t}{2}}(3 \cos(2t) + \frac{5}{4} \sin(2t)) $$ As time increases, the solution exhibits damped oscillatory behavior, with oscillations gradually decreasing in amplitude and approaching zero due to the damping effect of the exponential term.

Step-by-step solution

Solve the homogeneous differential equation

We have a second-order linear homogeneous differential equation, given by: $$ y^{\prime \prime} + y^{\prime} + 1.25y = 0 $$ To solve this, let's assume a solution of the form \(y(t) = e^{rt}\). Substituting this into the equation, we obtain the characteristic equation: $$ r^2 + r + 1.25 = 0 $$ We'll now solve for the roots of this quadratic equation.

Find the roots of the characteristic equation

To find the roots of the characteristic equation, we can use the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In our situation, \(a=1\), \(b=1\), and \(c=1.25\). Putting the values in the formula we get, $$ r = \frac{-1 \pm \sqrt{1 - 4 \cdot 1.25}}{2} = \frac{-1 \pm \sqrt{-4}}{2} $$ Since we have two complex roots, \(r_{1,2} = \frac{-1 \pm 2i}{2}\), our solution takes the form: $$ y(t) = e^{-\frac{t}{2}}(C_1 \cos(2t) + C_2 \sin(2t)) $$ Here, \(C_1\) and \(C_2\) are constants to be determined using the initial conditions.

Apply the initial conditions

The initial conditions are given by \(y(0)=3\) and \(y^{\prime}(0)=1\). Let's apply them to find the constants \(C_1\) and \(C_2\). First, using \(y(0)=3\), we get: $$ 3 = e^{0}(C_1 \cos(0) + C_2 \sin(0)) \Rightarrow C_1 = 3 $$ Now, we need to find \(y'(t)\). Differentiating \(y(t)\) with respect to \(t\), we get: $$ y^{\prime}(t) = -\frac{1}{2} e^{-\frac{t}{2}}(3 \cos(2t) + C_2 \sin(2t)) + e^{-\frac{t}{2}}(- 3 \sin(2t) + 2 C_2 \cos(2t)) $$ Now, apply the second initial condition \(y^{\prime}(0)=1\): $$ 1 = -\frac{1}{2} e^{0}(3 + 0) + e^{0}(0 + 2 C_2) \Rightarrow C_2 = \frac{5}{4} $$ So, the particular solution is: $$ y(t) = e^{-\frac{t}{2}}(3 \cos(2t) + \frac{5}{4} \sin(2t)) $$

Sketch the graph and describe the behavior

We observe that the damped oscillatory behavior of the solution is characterized by a decaying exponential envelope \(e^{-\frac{t}{2}}\), which damps the oscillations as \(t\) increases. The oscillations are sinusoidal, with a frequency of \(2\). To sketch the graph, plot the decaying exponential \(e^{-\frac{t}{2}}\) and oscillations \(3 \cos(2t) + \frac{5}{4} \sin(2t)\) together. The graph will show oscillatory behavior with an amplitude that decreases as \(t\) increases. In summary, as \(t\) increases, the solution oscillates with decreasing amplitude due to the damping effect of the exponential term, eventually approaching zero.