Problem 20 - Calculus: Early Transcendentals

Question

Solve the initial value problem. $$y^{\prime \prime}+6 y^{\prime}+18 y=\cos t-\sin t, y(0)=0, y^{\prime}(0)=2$$

Step-by-Step Solution

Answer

The final solution is obtained by combining the complementary solution and particular solution, then applying initial conditions to find arbitrary constants.

Step 1: Find the characteristic equation

The given differential equation is \( y'' + 6y' + 18y = \cos t - \sin t \). First, we solve the homogeneous part of this equation: \( y'' + 6y' + 18y = 0 \). The corresponding characteristic equation is \( r^2 + 6r + 18 = 0 \).

Step 2: Solve the characteristic equation

The characteristic equation \( r^2 + 6r + 18 = 0 \) is quadratic and can be solved using the quadratic formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1, b = 6, c = 18 \). Calculating the discriminant, we get:\( b^2 - 4ac = 36 - 72 = -36 \). Thus, the roots are complex: \[ r = \frac{-6 \pm \sqrt{-36}}{2} = -3 \pm 3i \].

Step 3: Write the complementary solution

Since the roots are complex \( -3 \pm 3i \), the complementary solution is:\( y_c(t) = e^{-3t}(C_1 \cos(3t) + C_2 \sin(3t)) \).

Step 4: Determine the particular solution

The non-homogeneous part of the differential equation is \( \cos t - \sin t \). We propose a particular solution of the form: \( y_p(t) = A \cos t + B \sin t \). Differentiating:\( y_p'(t) = -A \sin t + B \cos t \)\( y_p''(t) = -A \cos t - B \sin t \).Substitute these into the differential equation:\(-A \cos t - B \sin t + 6(-A \sin t + B \cos t) + 18(A \cos t + B \sin t) = \cos t - \sin t \).Simplify and equate coefficients to find \( A \) and \( B \).

Step 5: Solve for A and B

Continuing from Step 4, equate coefficients:For \( \cos t \):\( -A + 6B + 18A = 1 \rightarrow 17A + 6B = 1 \). For \( \sin t \):\( -B - 6A + 18B = -1 \rightarrow 12B - 6A = -1 \). Solve this system of equations:From the first equation, \( A = \frac{1 - 6B}{17} \).Substitute into the second equation to find \( B \) and then solve for \( A \).

Step 6: Find the complete solution

With values for \( A \) and \( B \) from Step 5, write the particular solution \( y_p(t) \). The complete solution is \( y(t) = y_c(t) + y_p(t) \).

Step 7: Apply initial conditions

The initial conditions are \( y(0) = 0 \) and \( y'(0) = 2 \). Substitute \( t = 0 \) into the complete solution \( y(t) \) and set it equal to 0, and also substitute into \( y'(t) \) and set it equal to 2. This will allow us to solve for the constants \( C_1 \) and \( C_2 \).

Step 8: Write the final solution

Once we have the values of \( C_1 \) and \( C_2 \), substitute them into \( y(t) = e^{-3t}(C_1 \cos(3t) + C_2 \sin(3t)) + y_p(t) \) to obtain the final solution for the initial value problem.

Key Concepts

Initial Value Problem

An initial value problem is a type of differential equation paired with specific conditions called initial conditions. These conditions specify the value of the unknown function and its derivatives at a particular point, usually at the start of the interval of interest. This helps to find a unique solution to the differential equation.

For example, in our problem, we have:

• The differential equation: \(y'' + 6y' + 18y = \cos t - \sin t \)
• The initial conditions: \(y(0) = 0\) and \(y'(0) = 2\)

The task is to find a function \(y(t)\) that satisfies both the differential equation and the given initial conditions. Initial value problems are particularly common in disciplines that involve motion over time, such as physics and engineering.

Characteristic Equation

The characteristic equation is a tool used when solving linear differential equations with constant coefficients. It is derived from the homogeneous part of the differential equation. This means we first ignore any non-homogeneous terms and focus on finding a general solution to the equation when it equals zero.

In our example, the differential equation was \(y'' + 6y' + 18y = 0\). From there, we derive the characteristic equation:

• \(r^2 + 6r + 18 = 0\)

Solving this quadratic equation helps us find the roots, which are crucial in constructing the complementary solution. The roots can be real or complex, and they determine the form of the solution.

Particular Solution

The particular solution adds the non-homogeneous part of the differential equation to our complete solution. This solution represents a specific function that satisfies the entire differential equation, including the non-homogeneous part.

For example, our non-homogeneous differential equation is \(y'' + 6y' + 18y = \cos t - \sin t\). After finding a complementary solution for the homogeneous part, we need to find a particular solution for the non-homogeneous term:

• We proposed: \(y_p(t) = A \cos t + B \sin t\)

Substituting this proposal into the differential equation allows us to solve for \(A\) and \(B\). By equating coefficients, we set up a system of equations to find the precise values of \(A\) and \(B\), ensuring \(y_p(t)\) satisfies the entire equation.

Complex Roots

Complex roots arise in the characteristic equation when the discriminant \( b^2 - 4ac \) is negative. These roots introduce trigonometric functions into the solution due to the presence of imaginary numbers.

In our situation, the roots of the characteristic equation \(r^2 + 6r + 18 = 0\) turned out to be complex:

• \(r = -3 \pm 3i\)

Complex roots result in solutions involving exponential and trigonometric functions. Specifically, the complementary solution takes the form:

• \(y_c(t) = e^{-3t}(C_1 \cos(3t) + C_2 \sin(3t))\)

The real part of the roots indicates an exponential decay, while the imaginary part introduces periodic behavior. This harmonizes into the general solution that combines steady decay and oscillation.