Problem 33 - Advanced Engineering Mathematics

Question

In Problems 33-38, solve the given differential equation subject to the indicated conditions. $$ y^{\prime \prime}-2 y^{\prime}+2 y=0, y(\pi / 2)=0, y(\pi)=-1 $$

Step-by-Step Solution

Answer

The particular solution is \( y(t) = e^{t - \pi} \cos t \).

Step 1: Identify the Type of Differential Equation

Recognize that the given equation \( y'' - 2y' + 2y = 0 \) is a second-order linear homogeneous differential equation with constant coefficients.

Step 2: Formulate the Characteristic Equation

To solve the differential equation, we first need to find the characteristic equation. This is done by assuming a solution of the form \( y = e^{rt} \) and substituting it into the differential equation. This yields the characteristic equation: \( r^2 - 2r + 2 = 0 \).

Step 3: Solve the Characteristic Equation for Roots

Solve the characteristic equation \( r^2 - 2r + 2 = 0 \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). For our equation, \( a = 1, b = -2, c = 2 \), so the roots are \( r = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 2}}{2} = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} \). Thus, \( r = 1 \pm i \).

Step 4: Write the General Solution

Since the roots are complex conjugates (\( 1 \pm i \)), the general solution of the differential equation is \( y(t) = e^{t}(C_1 \cos t + C_2 \sin t) \).

Step 5: Apply the Initial Conditions

Use the initial conditions to determine the constants \( C_1 \) and \( C_2 \). First condition: \( y(\frac{\pi}{2}) = 0 \). Substituting into the general solution, \( e^{\frac{\pi}{2}}(C_1 \cos \frac{\pi}{2} + C_2 \sin \frac{\pi}{2}) = 0 \), leading to \( C_2 e^{\frac{\pi}{2}} = 0 \), so \( C_2 = 0 \). This simplifies the solution to \( y(t) = C_1 e^{t} \cos t \).

Step 6: Apply the Second Initial Condition

Substitute the second initial condition \( y(\pi) = -1 \) into the simplified solution: \( e^{\pi}(C_1 \cos \pi) = -1 \). Since \( \cos \pi = -1 \), the equation becomes \( -C_1 e^\pi = -1 \), resulting in \( C_1 = e^{-\pi} \).

Step 7: Write the Particular Solution

Combine the values of the constants into the general solution to find the particular solution: \( y(t) = e^{t} e^{-\pi} \cos t = e^{t - \pi} \cos t \).

Key Concepts

Homogeneous Differential Equation

The differential equation you see here, \( y'' - 2y' + 2y = 0 \), is known as a homogeneous differential equation. But what does 'homogeneous' mean in this context? In simple terms, it indicates that every term in the equation involves the dependent variable \( y \) or its derivatives, without any standalone constants.
In a broader sense, homogeneous differential equations often have the form \( a_ny^{(n)} + a_{n-1}y^{(n-1)} + \dots + a_1y' + a_0y = 0 \). This form helps us in employing specific techniques for solutions, such as forming a characteristic equation. Conversely, non-homogeneous equations have an added term, \( f(x) \), which is independent of \( y \) or its derivatives.
Homogeneous differential equations are a stepping-stone to understanding more complex dynamics in mathematics. They pop-up in subjects that model wave behavior, heat transfer, and mechanical vibrations, making them universally applicable.

Characteristic Equation

To solve a homogeneous differential equation like \( y'' - 2y' + 2y = 0 \), you need to find its characteristic equation. This is a vital step in solving our differential equation.
The method involves constructing a characteristic equation by assuming a solution form \( y = e^{rt} \). Substitute \( y \) into the original equation and you'll derive the characteristic equation. For this exercise, the result is a quadratic equation: \( r^2 - 2r + 2 = 0 \).
The characteristic equation tells us about the roots (solutions) of our original differential equation. The types of roots you find, whether they're real or complex, dictate the form of the general solution, which captures all possible solutions of our differential equation.
By solving the characteristic equation at hand using the quadratic formula, we end up with complex roots, \( 1 \pm i \). This influences the structure of the general solution significantly.

Complex Roots

Complex roots arise when solving the characteristic equation, especially when the discriminant (the part under the square root in the quadratic formula) is negative. In our problem, the discriminate \( 4 - 8 = -4 \) indicates complex roots, resulting in \( r = 1 \pm i \).
When you have complex roots in the form \( \, \alpha \pm \beta i \, \), it impacts how you write the general solution to your differential equation. The form becomes \( y(t) = e^{\alpha t}(C_1 \cos \beta t + C_2 \sin \beta t) \).
In this exercise, since \( \alpha = 1 \) and \( \beta = 1 \), the solution is \( y(t) = e^t(C_1 \cos t + C_2 \sin t) \). Complex roots introduce oscillatory components, often seen in studies of physical systems like electrical circuits and mechanical structures oscillating under certain forces.

Initial Conditions

Initial conditions help us tailor the general solution of a differential equation to a specific situation. Without them, you'd have a family of solutions described by arbitrary constants \( C_1 \) and \( C_2 \). In this problem, you're given \( y(\frac{\pi}{2}) = 0 \) and \( y(\pi) = -1 \).
Using these conditions, you substitute into the general solution to solve for \( C_1 \) and \( C_2 \). For \( y(\frac{\pi}{2}) = 0 \), substituting gives \( C_2 = 0 \), simplifying our solution to \( y(t) = C_1 e^t \cos t \).
The next condition \( y(\pi) = -1 \) helps find \( C_1 \): \( -C_1 e^\pi = -1 \), leading to \( C_1 = e^{-\pi} \). Thus, applying initial conditions doesn't just solve for the constants; it pinpoints the specific path among the infinite generalized solutions, giving you a particular solution.