Problem 51 - Calculus 3

Question

For each of the following differential equations: a. Solve the initial value problem. b. [T] Use a graphing utility to graph the particular solution. $$ y^{\prime \prime}+5 y^{\prime}+15 y=0 \quad y(0)=-2, \quad y^{\prime}(0)=7 $$

Step-by-Step Solution

Answer

The solution is \( y(t) = e^{-\frac{5}{2}t}\left(-2 \cos\left(\frac{\sqrt{35}}{2}t\right) + \frac{4}{\sqrt{35}} \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \).

Step 1: Characteristic Equation

To solve the second order linear homogeneous differential equation \( y'' + 5y' + 15y = 0 \), we start with the characteristic equation given by \( r^2 + 5r + 15 = 0 \).

Step 2: Solve the Characteristic Equation

Solve for \( r \) using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], where \( a = 1, b = 5, \text{ and } c = 15 \). Calculate the discriminant: \( b^2 - 4ac = 25 - 60 = -35 \). This indicates complex roots: \( r = \frac{-5 \pm i\sqrt{35}}{2} \).

Step 3: Find the General Solution

Since the roots are complex, the general solution is \( y(t) = e^{-\frac{5}{2}t} \left(C_1 \cos\left(\frac{\sqrt{35}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \).

Step 4: Apply Initial Conditions

The initial conditions are \( y(0) = -2 \) and \( y'(0) = 7 \). Substitute \( t = 0 \) into the general solution to get \( -2 = C_1 \).Compute the derivative: \[ y'(t) = e^{-\frac{5}{2}t}\left(-\frac{5}{2}(C_1 \cos\left(\frac{\sqrt{35}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{35}}{2}t\right)) + C_2 \frac{\sqrt{35}}{2} \cos\left(\frac{\sqrt{35}}{2}t\right) - C_1 \frac{\sqrt{35}}{2} \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \]. Then use \( t = 0 \) in the derivative and solve \( 7 = -\frac{5}{2}C_1 + \frac{\sqrt{35}}{2}C_2 \) to get \( C_2 = \frac{14 + 5C_1}{\sqrt{35}} \approx \frac{14 + 5(-2)}{\sqrt{35}} \). Calculate \( C_2 \) resulting in \( C_2 = \frac{4}{\sqrt{35}} \approx 0.676 \).

Step 5: Write the Particular Solution

Substitute \( C_1 = -2 \) and \( C_2 \approx 0.676 \) into the general solution:\[ y(t) = e^{-\frac{5}{2}t}\left(-2 \cos\left(\frac{\sqrt{35}}{2}t\right) + \frac{4}{\sqrt{35}} \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \].

Step 6: Graph the Solution

Use a graphing utility to plot the particular solution. Input the equation:\( y(t) = e^{-\frac{5}{2}t} \left(-2 \cos\left(\frac{\sqrt{35}}{2}t\right) + \frac{4}{\sqrt{35}} \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \). The graph should reflect the initial conditions and exhibit exponential decay.

Key Concepts

Initial Value Problem

An initial value problem in differential equations not only involves finding the solution to the equation but also ensuring that this solution meets specific initial conditions. These initial conditions typically define the value of the function and its derivatives at a particular point, often when time, or \( t \), equals zero. In our exercise, we have a differential equation:

• \( y'' + 5y' + 15y = 0 \)

This equation comes with initial conditions:

• \( y(0) = -2 \)
• \( y'(0) = 7 \)

The goal is to find a solution \( y(t) \) that satisfies both the equation and these initial conditions. These conditions help us determine the specific constants in the general solution to fix the curve exactly to its required initial behavior.

Characteristic Equation

To solve a linear homogeneous differential equation, like the one in our example, we first need to derive the characteristic equation. This is done by assuming a solution of the form \( y = e^{rt} \) for the equation \( y'' + 5y' + 15y = 0 \). Substituting this form into the equation results in the characteristic equation:

• \( r^2 + 5r + 15 = 0 \)

This quadratic characteristic equation links the derivatives of the function with constant coefficients, which we need to solve in order to find the type of solutions (roots) we can expect. The roots of this equation will greatly determine the form of the general solution. In this case, we use the quadratic formula to solve for \( r \).

Complex Roots

Upon solving the characteristic equation for \( r \), we discover that the roots are complex. This happens when the discriminant \( b^2 - 4ac \) of the quadratic equation is negative. For our example equation:

• \( b^2 - 4ac = 25 - 60 = -35 \)
• Roots are calculated as \( r = \frac{-5 \pm i\sqrt{35}}{2} \)

Complex roots indicate the solution to the differential equation involves oscillatory components. The general solution will thus have exponential terms combined with sine and cosine functions:

• \( y(t) = e^{-rac{5}{2}t} \left(C_1 \cos\left(\frac{\sqrt{35}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \)

This reflects how the function both decays (due to the exponential term) and oscillates (due to the trigonometric terms).

General Solution

The general solution of a differential equation with complex roots typically comes from combining exponential, cosine, and sine functions. Using the complex roots found earlier:

• \( r = \frac{-5 \pm i\sqrt{35}}{2} \)

The general solution is:

• \( y(t) = e^{-rac{5}{2}t} \left(C_1 \cos\left(\frac{\sqrt{35}}{2}t\right) + C_2 \sin\left(\frac{\sqrt{35}}{2}t\right)\right) \)

Here, \( C_1 \) and \( C_2 \) are constants determined by the initial conditions. This solution provides a template to fit any similar differential equation with complex roots. By applying the initial value conditions \( y(0) = -2 \) and \( y'(0) = 7 \), we can find specific values for these constants and arrive at a particular solution tailored precisely to the problem at hand. The particular solution gives us a unique function that satisfies the initial value problem, reflecting both decay and oscillation patterns. This uniqueness is what differentiates general solutions from particular solutions.