Problem 83 - Calculus Volume 3

Question

Find the unique solution satisfying the differential equation and the initial conditions given, where \(y_{p}(x)\) is the particular solution. \(y^{\prime \prime}-5 y^{\prime}=e^{5 x}+8 e^{-5 x}\), \(y_{p}(x)=\frac{1}{5} x e^{5 x}+\frac{4}{25} e^{-5 x}, \quad y(0)=-2, \quad y^{\prime}(0)=0\)

Step-by-Step Solution

Answer

The unique solution is \(y(x) = -\frac{57}{25} + \frac{3}{25} e^{5x} + \frac{1}{5} x e^{5x} + \frac{4}{25} e^{-5x}\).

Step 1: Solve the Homogeneous Equation

The homogeneous equation associated with the differential equation is \(y'' - 5y' = 0\). To solve it, assume a solution of the form \(y_h = e^{rx}\), then substitute into the equation: \(r^2 e^{rx} - 5r e^{rx} = 0\). Factoring out \(e^{rx}\) gives \(e^{rx}(r^2 - 5r) = 0\). This leads to the characteristic equation \(r(r - 5) = 0\), which has roots \(r = 0\) and \(r = 5\). Hence, the general solution of the homogeneous equation is \(y_h(x) = C_1 + C_2 e^{5x}\).

Step 2: Write the General Solution

The general solution of the non-homogeneous equation is the sum of the homogeneous solution and the particular solution: \(y(x) = y_h(x) + y_p(x) = C_1 + C_2 e^{5x} + \frac{1}{5} x e^{5x} + \frac{4}{25} e^{-5x}\).

Step 3: Apply Initial Condition for \(y(0)\)

Substitute \(x = 0\) into the general solution and set it equal to the initial condition \(y(0) = -2\): \[y(0) = C_1 + C_2 e^{0} + \frac{1}{5}\cdot 0 \cdot e^{0} + \frac{4}{25} e^{0} = C_1 + C_2 + \frac{4}{25} = -2\].Solving for \(C_1 + C_2\), we get \(C_1 + C_2 = -2 - \frac{4}{25} = -\frac{54}{25}\).

Step 4: Find the Derivative \(y'(x)\)

Differentiate the general solution to find \(y'(x)\):\[y'(x) = C_2\cdot 5 e^{5x} + \frac{1}{5}(e^{5x} + 5x e^{5x}) - \frac{4}{25} \cdot 5 e^{-5x}\].Simplifying, we have \[y'(x) = 5C_2 e^{5x} + \frac{1}{5}e^{5x} + x e^{5x} - \frac{4}{5} e^{-5x}\].

Step 5: Apply Initial Condition for \(y'(0)\)

Substitute \(x = 0\) into the derivative and set it equal to the initial condition \(y'(0) = 0\):\[y'(0) = 5C_2 + \frac{1}{5} - \frac{4}{5} = 0\].Solving for \(C_2\), we find \(5C_2 = \frac{3}{5}\), so \(C_2 = \frac{3}{25}\).

Step 6: Solve for \(C_1\)

Plug \(C_2 = \frac{3}{25}\) back into the equation from Step 3: \(C_1 + C_2 = -\frac{54}{25}\). That gives:\[C_1 + \frac{3}{25} = -\frac{54}{25}\].Solving for \(C_1\), we find \(C_1 = -\frac{57}{25}\).

Step 7: Write the Unique Solution

Substitute the values of \(C_1\) and \(C_2\) into the general solution:\[y(x) = -\frac{57}{25} + \frac{3}{25} e^{5x} + \frac{1}{5} x e^{5x} + \frac{4}{25} e^{-5x}\].

Key Concepts

Homogeneous Equation

When faced with a differential equation like the one in this exercise, the first step is to solve the associated homogeneous equation. A homogeneous equation stems from removing any non-zero terms from the differential equation, essentially focusing on the part with the derivatives of the unknown function. In our example, this is given by removing the non-zero parts resulting in:

• The homogeneous form: \( y'' - 5y' = 0 \).
• We assume a solution of the form \( y_h = e^{rx} \).
• Substituting this into the homogeneous equation allows us to derive the characteristic equation.

Thus, finding solutions in the homogeneous scenario helps in forming the backbone for solving the non-homogeneous equation.

Particular Solution

The particular solution, noted as \( y_p(x) \), is crucial for addressing the non-zero part of the differential equation. It addresses the input functions or non-homogeneous terms specifically found on the right side of the equation.

• In our given exercise: \( y'' - 5y' = e^{5 x}+8 e^{-5 x} \), the inhomogeneous terms are \( e^{5 x} \) and \( 8 e^{-5 x} \).
• The provided particular solution is \( y_{p}(x)=\frac{1}{5} x e^{5 x}+\frac{4}{25} e^{-5 x} \).
• This is a specific solution to the non-homogeneous equation, capturing the behavior introduced by the non-homogeneous elements.

Particular solutions often require ``guessing'' a form of solution and refining it to fit the given equation.

Initial Conditions

Initial conditions are given values of the function and possibly its derivatives at a certain point. These conditions are integral to determining the specific solution from the general solution of a differential equation.

• For this problem, the initial conditions are \( y(0) = -2 \) and \( y'(0) = 0 \).
• By substituting these conditions into the general solution, we can solve for unknown constants, in this case, \( C_1 \) and \( C_2 \).
• They help in narrowing down from a broad set of possible solutions to a single, unique solution that satisfies both the differential equation and the initial conditions.

Initial conditions transform a family of solutions into a specific, meaningful solution relevant to the condition's context.

Characteristic Equation

The characteristic equation is derived during the process of finding the homogeneous solution. It emerges when you substitute \( y_h = e^{rx} \) into the homogeneous equation.

• Here, substituting \( y_h = e^{rx} \) into \( y'' - 5y' = 0 \) yields \( r^2 e^{rx} - 5r e^{rx} = 0 \).
• Simplifying, you factor \( e^{rx} \) out, resulting in \( r(r - 5) = 0 \).
• This leads to the characteristic equation \( r(r - 5) = 0 \), which gives roots of \( r = 0 \) and \( r = 5 \).
• The roots indicate the form of the homogeneous solution: \( y_h(x) = C_1 + C_2 e^{5x} \).
The characteristic equation facilitates understanding the behavior of the homogeneous equation's solutions and plays a fundamental role in forming the general solution.