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Magazine Chapter 4: Problem 68
Solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}+5 y^{\prime}-6 y=10 e^{2 x}, \quad y(0)=1, y^{\prime}(0)=1 $$
Short Answer
Expert verified
The solution is \( y(x) = \frac{3}{8} e^x - \frac{5}{8} e^{-6x} + \frac{5}{4} e^{2x} \).
Step by step solution
01
Solve the Homogeneous Equation
Start by solving the associated homogeneous equation: \[ y'' + 5y' - 6y = 0 \]Assume a solution of the form \( y = e^{rx} \). Substitute into the homogeneous equation to get the characteristic equation:\[ r^2 + 5r - 6 = 0 \]Solve this quadratic equation using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1, b = 5, c = -6 \). Calculate the roots:\[ r = \frac{-5 \pm \sqrt{5^2 + 24}}{2} \]\[ r = \frac{-5 \pm 7}{2} \]This gives roots \( r = 1 \) and \( r = -6 \). Thus, the general solution of the homogeneous equation is:\[ y_h = C_1 e^x + C_2 e^{-6x} \]
02
Solve the Non-Homogeneous Equation
The non-homogeneous part of the differential equation is \( 10e^{2x} \). We'll use the method of undetermined coefficients to find a particular solution.Assume a particular solution of the form \( y_p = Ae^{2x} \). Substitute \( y_p \), \( y'_p = 2Ae^{2x} \), and \( y''_p = 4Ae^{2x} \) into the original equation:\[ 4Ae^{2x} + 5(2Ae^{2x}) - 6Ae^{2x} = 10e^{2x} \]Simplify to:\[ 4A + 10A - 6A = 10 \]\[ 8A = 10 \]\[ A = \frac{5}{4} \]Thus, the particular solution is \( y_p = \frac{5}{4}e^{2x} \).
03
Form the General Solution
The general solution of the differential equation is the sum of the homogeneous and particular solutions:\[ y(x) = C_1 e^x + C_2 e^{-6x} + \frac{5}{4}e^{2x} \]
04
Apply Initial Conditions
Use the initial conditions to determine \( C_1 \) and \( C_2 \).First, use \( y(0) = 1 \):\[ y(0) = C_1 e^0 + C_2 e^0 + \frac{5}{4}e^0 = 1 \]\[ C_1 + C_2 + \frac{5}{4} = 1 \]\[ C_1 + C_2 = \frac{-1}{4} \] (Equation A)Now, use the initial condition \( y'(0) = 1 \). First calculate \( y'(x) \):\[ y'(x) = C_1 e^x - 6C_2 e^{-6x} + \frac{5}{2}e^{2x} \]\[ y'(0) = C_1 - 6C_2 + \frac{5}{2} = 1 \]\[ C_1 - 6C_2 = \frac{-3}{2} \] (Equation B)Solve Equations A and B simultaneously to find \( C_1 \) and \( C_2 \):Adding Equation A and Equation B:\[ 2C_1 - 5C_2 = -2 \]Solve to get \( C_1 = \frac{3}{8} \) and \( C_2 = \frac{-5}{8} \).
05
Write the Final Solution
Substitute the constants back into the general solution:\[ y(x) = \frac{3}{8} e^x - \frac{5}{8} e^{-6x} + \frac{5}{4} e^{2x} \]This is the solution that satisfies the given differential equation and initial conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Conditions
To fully solve a differential equation, one key element is the initial conditions. These are specific values given for the solution and its derivatives at a certain point, often designated as when the independent variable equals zero. In our problem:
- The initial conditions are given as \( y(0) = 1 \) and \( y'(0) = 1 \). These conditions provide essential information that allows us to find the unique solution out of the infinite possibilities that satisfy the differential equation.
- Initial conditions are used to solve for the constants in the general solution of a differential equation. This is crucial because without these constants, the solution would remain in a form that represents a family of functions.
- In this exercise, after obtaining the general solution, the initial conditions were applied to determine the particular values of \( C_1 \) and \( C_2 \), which in turn ensures that the solution fits the specific requirements.
Homogeneous Equation
The homogeneous equation is a special form of a differential equation set to zero. It provides the baseline solution from which all other solutions can be derived. In mathematical terms, it looks like:
- For the given problem, the homogeneous equation is \( y'' + 5y' - 6y = 0 \).
- The homogeneous equation is often solved by assuming a solution of the form \( y = e^{rx} \). This turns our differential equation into a characteristic equation, which simplifies our work greatly.
- The roots of the characteristic equation determine the form of the solution for the homogeneous equation, whether it includes exponential functions, oscillatory functions, or a combination.
Particular Solution
The particular solution of a differential equation accounts for non-homogeneous terms, which are the terms that aren't zero in the equation. It considers the inhomogeneity in the differential equation and is often sought after the homogeneous solution.
- For our differential equation, we have a non-homogeneous term \( 10e^{2x} \), and we assume a particular solution of the form \( y_p = Ae^{2x} \).
- To find \( A \), this assumed particular solution is substituted back into the original differential equation.
- The coefficients are equated to find the exact value, giving us a particular solution that satisfies the differential equation entirely.
Method of Undetermined Coefficients
The method of undetermined coefficients is an effective technique used to find particular solutions for linear non-homogeneous differential equations with constant coefficients. It involves guessing a form based on the non-homogeneous term and is refined to find the correct coefficients.
- In this problem, the non-homogeneous term \( 10e^{2x} \) suggests a particular solution form of \( Ae^{2x} \).
- The next step involves substituting this guessed solution form into the differential equation and solving for the coefficients, which here is \( A \).
- This method is especially convenient for differential equations with right-hand sides comprised of polynomial, exponential, or trigonometric functions because the guessed form is consistent and predictable.
