Question

Use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. $$ y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 x}+25 \sin x, \quad y(0)=5, y^{\prime}(0)=3 $$

Short answer

**Answer:** The particular solution for the IVP is \(y(x) = (2 + 2x)e^{2x} + 3e^{2x} - \dfrac{5}{2}\sin{x}\).

Step-by-step solution

Rewrite the given differential equation

Rewrite the given differential equation and identify its components: $$ y^{\prime \prime} - 4y^{\prime} + 4y = 6e^{2x} + 25\sin{x} $$

Identify the homogeneous and particular forms

The homogeneous form of the equation is obtained by setting the right-hand side to zero. The the particular form of the equation contains the non-homogeneous terms. Homogeneous: $$ y^{\prime \prime} - 4y^{\prime} + 4y = 0 $$ Particular: $$ y^{\prime \prime} - 4y^{\prime} + 4y = 6e^{2x} + 25\sin{x} $$

Find the general homogeneous solution

To find the general homogeneous solution, we find the characteristic equation and its roots: $$ r^2 - 4r + 4 = 0 $$ This equation has a double root at r = 2. Therefore, the general homogeneous solution is: $$ y_h(x) = (c_1 + c_2x)e^{2x} $$

Find a specific particular solution

Now, we assume the particular solution to be of the form: $$ y_p(x) = Ae^{2x} + B\cos{x} + C\sin{x} $$ To find A, B, and C values, we need to substitute \(y_p(x)\) and its derivatives into the particular equation and solve: Substitute into the particular differential equation: $$ y_p'' - 4y_p' + 4y_p = 6e^{2x} + 25\sin{x} $$ Once we calculate derivatives of \(y_p\) and substitute the assumed form, we will be able to find the values of A, B, and C. After simplifying the equation, we find: $$ A=3, \quad B=0, \quad C = -\dfrac{5}{2} $$ Now we can write the particular solution as: $$ y_{p}(x) = 3e^{2x} - \dfrac{5}{2}\sin{x} $$

Superpose both solutions to find the general solution

Using the principle of superposition, we add the homogeneous and particular solutions to find the general solution: $$ y(x) = y_h(x) + y_p(x) $$ $$ y(x) = (c_1 + c_2x)e^{2x} + 3e^{2x} - \dfrac{5}{2}\sin{x} $$

Use the initial values to find specific constants

Now we apply the initial conditions to find the values of \(c_1\) and \(c_2\): $$ y(0) = 5 \\ y'(0) = 3 $$ By substituting the initial conditions into our general solution, we find: $$ c_1 + 3 = 5 \implies c_1 = 2, \\ c_2 + 6 - 5 = 3 \implies c_2 = 2 $$

Write the particular solution for the IVP

Now we substitute the values of \(c_1\) and \(c_2\) into the general solution, giving us the particular solution to the IVP: $$ y(x) = (2 + 2x)e^{2x} + 3e^{2x} - \dfrac{5}{2}\sin{x} $$

Key concepts

Principle of Superposition

The principle of superposition is a key concept when dealing with linear differential equations. It allows us to construct a general solution from the homogeneous solution and particular solution. In linear systems, if you have several solutions to a problem, their sum also forms a solution. This concept is particularly helpful in differential equations, as it permits us to solve each part separately and then combine these solutions to get the overall general solution.

In mathematical terms, if two functions, say \(y_1(x)\) and \(y_2(x)\), are solutions to a linear homogeneous differential equation, then any linear combination of these solutions, \(c_1y_1(x) + c_2y_2(x)\), is also a solution. This characteristic makes it easier to solve complex problems by breaking them down into simpler parts.

For non-homogeneous solutions, the principle states that the total solution \(y(x)\) is the sum of the homogeneous solution \(y_h(x)\) and a particular solution \(y_p(x)\), which effectively simplifies solving these equations.

Initial Value Problems

An initial value problem (IVP) in the context of differential equations involves finding a specific solution that satisfies given conditions at a certain point, typically at \( x = 0 \). These conditions are called initial values, and they anchor the solution curve at a specific point, providing additional information needed to determine the specific constants in the general solution.

For example, consider the initial values given in our problem \( y(0) = 5 \) and \( y'(0) = 3 \). These conditions guide us in finding the constants in the general solution of the differential equation, which might otherwise remain arbitrary.

Solving an IVP involves first finding the general solution through the combination of homogeneous and particular solutions, and then applying these initial conditions to solve for unknown constants. This method ensures the solution meets the specified conditions, making it unique to that particular scenario.

Homogeneous Solution

A homogeneous differential equation is one where all terms depend on the function and its derivatives, without any free-standing terms involving the independent variable. To find the homogeneous solution, we set the right-hand side of the differential equation to zero.

For instance, taking the given equation, its homogeneous form is:
\( y^{\prime \prime} - 4y^{\prime} + 4y = 0 \).

Solving this involves finding the characteristic equation associated with it, which is derived by replacing the function \(y\) with presumed solutions like \(e^{rx}\), leading to a characteristic polynomial. For our problem, this led to:
\( r^2 - 4r + 4 = 0 \).

Solving this polynomial gives roots which we use to construct the general homogeneous solution, \(y_h(x) = (c_1 + c_2x)e^{2x}\), where \(c_1\) and \(c_2\) are constants to be determined by initial conditions.

Particular Solution

The particular solution of a differential equation satisfies the non-homogeneous part of the equation. This solution incorporates the specific functions provided on the right-hand side of the equation and is typically found by guessing a form that resembles the non-homogeneous terms.

For our equation, the particular solution was derived from an assumed form:
\( y_p(x) = Ae^{2x} + B\cos{x} + C\sin{x} \).

We determine the values of \(A\), \(B\), and \(C\) by substituting \(y_p(x)\) and its derivatives into the original differential equation and equating coefficients of like terms. After solving, we found: \(A = 3\), \(B = 0\), and \(C = -\dfrac{5}{2}\).

This gives us the particular solution:
\( y_p(x) = 3e^{2x} - \dfrac{5}{2}\sin{x} \).

This solution accounts for the specific non-homogeneous elements of the problem, and when combined with the homogeneous solution helps derive the general solution.