Question

In Exercises \(13-17\) solve the initial value problem. $$ 4 y^{\prime \prime}-12 y^{\prime}+9 y=0, \quad y(0)=3, \quad y^{\prime}(0)=\frac{5}{2} $$

Short answer

Answer: The general solution is \(y(x) = 3 e^{\frac{3}{2} x} - x e^{\frac{3}{2} x}\).

Step-by-step solution

Find the complementary function

To find the complementary function, we first write the characteristic equation for the given differential equation: $$ 4r^2 - 12r + 9 = 0 $$ This is a quadratic equation, which can be factored as: $$ (2r - 3)(2r - 3) = 0 $$ The equation has a double real root: \(r = \frac{3}{2}\). Therefore, the complementary function (CF) is given by: $$ y_c(x) = C_1 e^{\frac{3}{2} x} + C_2 x e^{\frac{3}{2} x} $$ where \(C_1\) and \(C_2\) are constants to be determined.

Find the particular solution using the initial conditions

The initial conditions are: $$ y(0) = 3, \quad y'(0) = \frac{5}{2} $$ Applying the first initial condition, we have: $$ 3 = y(0) = C_1 e^{0} + C_2 (0) e^{0} \implies C_1 = 3 $$ Now, we need to find \(y'(x)\) to use the second initial condition: $$ y'(x) = \frac{3}{2} C_1 e^{\frac{3}{2} x} + C_2 e^{\frac{3}{2} x} + \frac{3}{2} C_2 x e^{\frac{3}{2} x} $$ Applying the second initial condition: $$ \frac{5}{2} = y'(0) = \frac{3}{2} (3) e^{0} + C_2 e^{0} \implies C_2 = -1 $$

Write the general solution

Now that we have the values for \(C_1\) and \(C_2\), we can write the general solution of the IVP as: $$ y(x) = 3 e^{\frac{3}{2} x} - x e^{\frac{3}{2} x} $$ And that's the final solution for the given initial value problem.

Key concepts

Initial Value Problem

An initial value problem in differential equations involves finding a solution to a differential equation that also satisfies certain specified initial conditions. In simple terms, it's like solving a mystery where you know the equation some object must follow and its starting point. In our exercise, the initial value conditions are given as \( y(0) = 3 \) and \( y'(0) = \frac{5}{2} \).
These conditions help determine the specific solution to the differential equation \( 4 y'' - 12 y' + 9 y = 0 \).

• \( y(0) = 3 \): This defines the value of the solution function at \( x = 0 \).
• \( y'(0) = \frac{5}{2} \): This defines the slope of the tangent to the solution at \( x = 0 \).

Having these initial conditions allows us to find the constants in the solution, ensuring it fits perfectly within the "initial scenario" or "state" described by the problem.

Characteristic Equation

The characteristic equation is derived from a linear differential equation and is key in determining its solutions. For a differential equation like \( 4 y'' - 12 y' + 9 y = 0 \), we start by assuming solutions of the form \( y = e^{rt} \), where \( r \) will be found.
This leads us to create the characteristic equation by inserting this assumption into the original differential equation, which results in:\[4r^2 - 12r + 9 = 0\]Solving this equation gives values for \( r \), which are used to construct the solutions to the original differential equation. The characteristic equation essentially boils down the problem to solving algebraic equations like the quadratic one we derived.

Complementary Function

The complementary function represents the general solution to the homogeneous part of a differential equation. In our exercise, after finding the characteristic equation \( 4r^2 - 12r + 9 = 0 \) and its roots, we can determine the form of the complementary function.
In our problem, the characteristic equation has a double root of \( r = \frac{3}{2} \), meaning both roots are the same. This affects the form of the complementary function: \[y_c(x) = C_1 e^{\frac{3}{2} x} + C_2 x e^{\frac{3}{2} x}\]
This solution form arises because when you have repeated roots, you need to multiply one of the terms by \( x \) to ensure the solution isn't identical and linearly independent. The constants \( C_1 \) and \( C_2 \) are later determined using initial conditions.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two, typically in the form \( ax^2 + bx + c = 0 \). Solving these equations can be done using factoring, the quadratic formula, or completing the square.
In the context of our problem, the characteristic equation \( 4r^2 - 12r + 9 = 0 \) is a quadratic equation. To solve it, we factor it into \((2r - 3)^2 = 0\), revealing that \( r = \frac{3}{2} \) is a double root.
This means that both solutions to the quadratic equation are the same, which implies a specific structure in the general solution to the differential equation. Quadratic equations often pop up in the study of differential equations as they help find solutions efficiently and clearly.