Chapter 4: Problem 10
In Problems 1-24 find a second solution of each differential equation. Use reduction of order or formula (5) as instructed. Assume an appropriate interval of validity. $$ 6 y^{\prime \prime}+y^{\prime}-y=0 ; \quad y_{1}=e^{x^{2}} $$
Short Answer
Expert verified
The second solution is \( y_2 = x e^{x^2} \).
Step by step solution
01
Recognize the context and method to use
We are given that the differential equation is \( 6y'' + y' - y = 0 \) with a solution \( y_1 = e^{x^2} \). The problem instructs us to find a second solution using the method of reduction of order. This is applicable because we know one solution, \( y_1 \).
02
Apply reduction of order
In reduction of order, we assume the second solution \( y_2 \) has the form \( y_2 = v(x)y_1 \), where \( v(x) \) is a function to be determined. Hence, \( y_2 = v(x)e^{x^2} \).
03
Differentiate assumed solution
Find the derivatives of \( y_2 = v(x) e^{x^2} \):\( y_2' = v'(x) e^{x^2} + 2x v(x) e^{x^2} \)\( y_2'' = v''(x) e^{x^2} + 4x v'(x) e^{x^2} + (4x^2 + 2) v(x) e^{x^2} \).
04
Substitute back into the differential equation
Substitute the derivatives back into the differential equation:\( 6(v'' e^{x^2} + 4x v' e^{x^2} + (4x^2+2)v e^{x^2}) + (v'e^{x^2} + 2xv e^{x^2}) - ve^{x^2} = 0 \).
05
Simplify and solve for \( v(x) \)
Factor out \( e^{x^2} \):\[ 6v'' + (24x+1) v' + (24x^2+12x+1) v = 0 \]Since \( y_1 = e^{x^2} \) is a solution of the homogeneous equation, \( y_2 \) must fulfill:\( v'' = 0 \). Integrate to find \( v(x) \):First integral: \( v'(x) = C_1 \)Second integral: \( v(x) = C_1 x + C_2 \).
06
Form the second solution \( y_2 \)
Since \( v(x) = C_1 x + C_2 \), substitute back to form \( y_2 \):\( y_2 = (C_1 x + C_2) e^{x^2} \).To avoid duplicating the first solution and to find a distinctly different second solution, we set \( C_2 = 0 \), giving:\( y_2 = C_1 x e^{x^2} \).Any multiple of this is also a solution, so \( y_2 = x e^{x^2} \) is a valid second solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reduction of Order
The reduction of order is a technique used to find a second solution to a linear homogeneous differential equation when one solution is already known. The importance of this method lies in its ability to produce a second, linearly independent solution. This is critical for forming the general solution of a differential equation.
Here’s how it works:
By solving this simpler equation, we can confidently proceed to find the second solution, confirming the independence from the given first solution.
Here’s how it works:
- Given a solution, say \( y_1 \), we assume the second solution, \( y_2 \), has the form \( y_2 = v(x) y_1 \). Here, \( v(x) \) is an unknown function that we need to determine.
- We then differentiate this assumed solution and substitute back into the original differential equation to solve for \( v(x) \).
- Through integration, we can find \( v(x) \), thus constructing our second solution \( y_2 \).
By solving this simpler equation, we can confidently proceed to find the second solution, confirming the independence from the given first solution.
Homogeneous Differential Equations
A homogeneous differential equation is one in which every term is a multiple of the dependent variable or its derivatives. Essentially, it can be written in the form: \[ a_n(x) y^{(n)} + a_{n-1}(x) y^{(n-1)} + ... + a_0(x) y = 0 \] where each term involves the function \(y\) or its derivatives.
This form is crucial because it suggests the equation has solutions of the form \(e^{rx}\), allowing for systematic solution methods such as reduction of order or undetermined coefficients.
In the solution process:
This form is crucial because it suggests the equation has solutions of the form \(e^{rx}\), allowing for systematic solution methods such as reduction of order or undetermined coefficients.
In the solution process:
- The first solution is often found by inspection or using standard methods like characteristic equations.
- The equation's adherent homogeneity assures that any linear combination of solutions is also a solution.
Second Solution
The search for a 'second solution' is crucial in solving linear homogeneous differential equations. Given a first solution, you require another to establish the general solution, typically in the form: \[ y = C_1 y_1 + C_2 y_2 \] where \(C_1\) and \(C_2\) are constants.
Having two linearly independent solutions ensures that all possible solutions can be represented and explored.
To identify a second solution:
Having two linearly independent solutions ensures that all possible solutions can be represented and explored.
To identify a second solution:
- Use reduction of order when one solution is given, as demonstrated. Assume the form \(y_2 = v(x)y_1\) and derive \(v(x)\) through integration.
- Ensure the second solution is linearly independent. This means \(y_2\) should not be a constant multiple of \(y_1\).
Method of Undetermined Coefficients
The method of undetermined coefficients is a powerful tool used primarily to find a particular solution to non-homogeneous linear differential equations. However, it is mentioned here as it complements the homogeneous case.
Here's a brief outline of how this method is applied:
It sets the stage for dealing with a broader class of problems, thus empowering a student to manage both homogeneous and non-homogeneous differential equations effectively.
Here's a brief outline of how this method is applied:
- Assume a form for the particular solution based on the non-homogeneous term. This involves guessing functions with undetermined coefficients.
- Substitute this assumed solution into the differential equation.
- Solve for the coefficients by ensuring the equation holds for all x.
It sets the stage for dealing with a broader class of problems, thus empowering a student to manage both homogeneous and non-homogeneous differential equations effectively.