Question

Solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing \(t.\) \(9 y^{\prime \prime}+6 y^{\prime}+82 y=0, \quad y(0)=-1, \quad y^{\prime}(0)=2\)

Short answer

Short Answer: The particular solution of the given initial value problem is \(y(t)=e^{-\frac{1}{3}t}[-\cos(\frac{\sqrt{696}}{18}t)+\frac{9}{2}\sin(\frac{\sqrt{696}}{18}t)]\). Its graph represents damped oscillations, where the amplitude of the oscillations decreases as time goes on, eventually approaching zero. The solution indicates a stable equilibrium as time increases.

Step-by-step solution

Identify the given differential equation and initial conditions

The given differential equation is \(9y^{\prime\prime}+6y^{\prime}+82y=0\) with initial conditions \(y(0)=-1\) and \(y^{\prime}(0)=2\).

Determine the characteristic equation

The characteristic equation is found by replacing \(y^{\prime\prime}\) with \(r^2,\) \(y^{\prime}\) with \(r,\) and \(y\) with \(1\). This gives the characteristic equation: \(9r^2+6r+82=0\).

Solve the characteristic equation

To solve the quadratic equation \(9r^2+6r+82=0\), we use the quadratic formula: \(r=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\). In this case, \(a=9\), \(b=6\), and \(c=82\). So, the roots are: \(r=\frac{-6\pm\sqrt{6^2-4(9)(82)}}{2(9)}\). Since the discriminant \(b^2-4ac\) is negative, we'll have complex roots. Therefore, the general solution of the differential equation will be in the form: \(y(t)=e^{\alpha t}[A\cos(\beta t)+B\sin(\beta t)]\), where \(\alpha\) is the real part of the roots and \(\beta\) is the imaginary part of the roots.

Find the general solution of the differential equation

We obtain the real part \(\alpha\) and the imaginary part \(\beta\) from our roots: \(\alpha=\frac{-6}{2(9)}=-\frac{1}{3}\), and \(\beta=\frac{\sqrt{|6^2-4(9)(82)|}}{2(9)}=\frac{\sqrt{696}}{18}\). Now, we substitute these values into the general solution form: \(y(t)=e^{-\frac{1}{3}t}[A\cos(\frac{\sqrt{696}}{18}t)+B\sin(\frac{\sqrt{696}}{18}t)]\).

Apply the initial conditions to obtain the particular solution

Apply the initial condition \(y(0)=-1\): \(-1=e^{0}[A\cos(0)+B\sin(0)]\) which simplifies to \(A=-1\). Apply the initial condition \(y^{\prime}(0)=2\): First, find the derivative \(y^{\prime}(t)\): \(y^{\prime}(t)=-\frac{1}{3}e^{-\frac{1}{3}t}[-A\cos(\frac{\sqrt{696}}{18}t)-B\sin(\frac{\sqrt{696}}{18}t)]+\frac{\sqrt{696}}{18}e^{-\frac{1}{3}t}[-A\sin(\frac{\sqrt{696}}{18}t)+B\cos(\frac{\sqrt{696}}{18}t)]\) Substitute \(A=-1\) and evaluate at \(t=0\): \(2=-\frac{1}{3}e^{0}[-(-1)\cos(0)-B\sin(0)]+\frac{\sqrt{696}}{18}e^{0}[-(-1)\sin(0)+B\cos(0)]\) which simplifies to \(B=\frac{9}{2}\). The particular solution is: \(y(t)=e^{-\frac{1}{3}t}[-\cos(\frac{\sqrt{696}}{18}t)+\frac{9}{2}\sin(\frac{\sqrt{696}}{18}t)]\).

Sketch the graph of the particular solution and describe its behavior for increasing \(t\)

The graph of particular solution represents a damped oscillation, due to the presence of the exponential term \(e^{-\frac{1}{3}t}\) and the trigonometric functions. As \(t\) increases, the amplitude of the oscillations will decrease, eventually approaching zero. This means that the solution approaches a stable equilibrium as time goes on.