Problem 19
Question
The indicated function \(y_{1}(x)\) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution \(y_{2}(x)\) of the homogeneous equation and a particular solution \(y_{p}(x)\) of the given non-homogeneous equation. $$y^{\prime \prime}-3 y^{\prime}+2 y=5 e^{3 x} ; \quad y_{1}=e^{x}$$
Step-by-Step Solution
Verified Answer
The second solution is \(y_2(x) = xe^x\), and a particular solution is \(y_p(x) = \frac{5}{2}e^{3x}\).
1Step 1: Identify the Homogeneous Equation
The given differential equation is \( y'' - 3y' + 2y = 5e^{3x} \). The associated homogeneous equation is \( y'' - 3y' + 2y = 0 \).
2Step 2: Verify Given Solution
Check that \( y_1(x) = e^x \) is a solution of the homogeneous equation by substituting it into \( y'' - 3y' + 2y = 0 \). Calculate the derivatives: \( y'_1 = e^x \) and \( y''_1 = e^x \). Substitute to see \( e^x - 3e^x + 2e^x = 0 \), confirming it is a solution.
3Step 3: Apply Reduction of Order
Propose \( y_2(x) = v(x)e^x \) where \( v(x) \) is an unknown function. Calculate derivatives: \( y'_2 = v'e^x + ve^x \) and \( y''_2 = v''e^x + 2v'e^x + ve^x \). Substitute into the homogeneous equation to simplify.
4Step 4: Solve for the Second Solution
Substitute and simplify the terms in the homogeneous equation: \( v''e^x + 2v'e^x + ve^x - 3v'e^x - 3ve^x + 2ve^x = 0 \). This simplifies to \( v''e^x = 0 \). From this, we deduce \( v'' = 0 \), leading to \( v' = C_1 \), and \( v(x) = C_1x + C_2 \). So, \( y_2(x) = x e^x \).
5Step 5: Find Particular Solution Using Undetermined Coefficients
For the non-homogeneous equation, guess a particular solution of the form \( y_p(x) = Ae^{3x} \). Substitute into the differential equation: \( y''_p = 9Ae^{3x} \), \( y'_p = 3Ae^{3x} \). Substitute into the equation to obtain: \( 9Ae^{3x} - 9Ae^{3x} + 2Ae^{3x} = 5e^{3x} \) which gives \( 2A = 5 \). Therefore, \( A = \frac{5}{2} \) and \( y_p(x) = \frac{5}{2}e^{3x} \).
6Step 6: Combine Solutions
The general solution of the non-homogeneous differential equation is a linear combination of the solutions of the homogeneous equation and the particular solution: \( y(x) = C_1e^x + C_2xe^x + \frac{5}{2}e^{3x} \), where \( C_1 \) and \( C_2 \) are constants.
Key Concepts
Homogeneous EquationReduction of OrderParticular Solution
Homogeneous Equation
A homogeneous differential equation is one where the right-hand side equals zero. For example, our starting equation is \( y^{\prime\prime} - 3y^{\prime} + 2y = 0 \). It only involves the function \( y \) and its derivatives, without any external factors or "forcing terms." This means that everything in the equation depends on the solution \( y \) itself.
To identify that an equation is homogeneous, look for the zero on the right side of the equation. This is crucial as it indicates that the equation's form allows us to apply specific methods like the reduction of order to find its solutions. Understanding this gives you a solid base since solutions to homogeneous equations form the foundation for solving non-homogeneous equations.
Solving a homogeneous equation typically involves finding functions, like \( e^x \), which when substituted back into the equation reduce everything to zero. This characteristic zero balance differentiates homogeneous equations from their non-homogeneous counterparts.
To identify that an equation is homogeneous, look for the zero on the right side of the equation. This is crucial as it indicates that the equation's form allows us to apply specific methods like the reduction of order to find its solutions. Understanding this gives you a solid base since solutions to homogeneous equations form the foundation for solving non-homogeneous equations.
Solving a homogeneous equation typically involves finding functions, like \( e^x \), which when substituted back into the equation reduce everything to zero. This characteristic zero balance differentiates homogeneous equations from their non-homogeneous counterparts.
Reduction of Order
The method of reduction of order is handy when you already know one solution to a second-order homogeneous differential equation. This known solution helps find a necessary second solution.
For our equation, one solution is \( y_1(x) = e^x \). To find another, assume \( y_2(x) = v(x) e^x \), where \( v(x) \) is a new unknown function. The idea is that multiplying \( v(x) \) with the known solution preserves the integrity of the solution for the differential equation.
After substituting \( y_2 \) into the equation and simplifying, you arrive at \( v'' = 0 \). Solving \( v'' = 0 \) involves basic integration, leading to \( v'(x) = C_1 \) and then \( v(x) = C_1 x + C_2 \). Thus, \( y_2(x) = C_1 x e^x + C_2 e^x \), typically simplified as \( y_2(x) = x e^x \) when looking for a linearly independent solution in conjunction with \( y_1 \).
Understanding reduction of order equips us to efficiently extend known solutions, making it a powerful tool in the differential equations toolkit.
For our equation, one solution is \( y_1(x) = e^x \). To find another, assume \( y_2(x) = v(x) e^x \), where \( v(x) \) is a new unknown function. The idea is that multiplying \( v(x) \) with the known solution preserves the integrity of the solution for the differential equation.
After substituting \( y_2 \) into the equation and simplifying, you arrive at \( v'' = 0 \). Solving \( v'' = 0 \) involves basic integration, leading to \( v'(x) = C_1 \) and then \( v(x) = C_1 x + C_2 \). Thus, \( y_2(x) = C_1 x e^x + C_2 e^x \), typically simplified as \( y_2(x) = x e^x \) when looking for a linearly independent solution in conjunction with \( y_1 \).
Understanding reduction of order equips us to efficiently extend known solutions, making it a powerful tool in the differential equations toolkit.
Particular Solution
In contrast to homogeneous equations, non-homogeneous differential equations include a non-zero term, the "forcing term." For instance, \( y^{\prime\prime} - 3y^{\prime} + 2y = 5e^{3x} \). The solution to such equations combines solutions to the homogeneous equation with a particular solution that depends on this forcing term.
To find a particular solution, try a function that resembles the non-zero part of your equation. In this case, assume a solution like \( y_p(x) = A e^{3x} \). Substituting \( y_p \) into the equation involves calculating derivatives \( y'_p = 3A e^{3x} \) and \( y''_p = 9A e^{3x} \), and simplifying gives \( 2A = 5 \).
Solving for \( A \), we find \( A = \frac{5}{2} \), leading to \( y_p(x) = \frac{5}{2} e^{3x} \). This particular solution accounts for the non-zero external component, ensuring that when paired with solutions from the homogeneous equation, the entire system is satisfied. This synergy between general and particular solutions in differential equations showcases both elegant mathematical structure and practical application potential.
To find a particular solution, try a function that resembles the non-zero part of your equation. In this case, assume a solution like \( y_p(x) = A e^{3x} \). Substituting \( y_p \) into the equation involves calculating derivatives \( y'_p = 3A e^{3x} \) and \( y''_p = 9A e^{3x} \), and simplifying gives \( 2A = 5 \).
Solving for \( A \), we find \( A = \frac{5}{2} \), leading to \( y_p(x) = \frac{5}{2} e^{3x} \). This particular solution accounts for the non-zero external component, ensuring that when paired with solutions from the homogeneous equation, the entire system is satisfied. This synergy between general and particular solutions in differential equations showcases both elegant mathematical structure and practical application potential.
Other exercises in this chapter
Problem 19
Solve each differential equation by variation of parameters, subject to the initial conditions \(y(0)=1, y^{\prime}(0)=0\). $$4 y^{\prime \prime}-y=x e^{x / 2}$
View solution Problem 19
Find a linear differential operator that annihilates the given function. $$\cos 2 x$$
View solution Problem 19
Find the general solution of the given higher order differential equation. $$\frac{d^{3} u}{d t^{3}}+\frac{d^{2} u}{d t^{2}}-2 u=0$$
View solution Problem 19
Determine whether the given set of functions is linearly independent on the interval \((-\infty, \infty)\). $$f_{1}(x)=x, \quad f_{2}(x)=x-1, \quad f_{3}(x)=x+3
View solution