Chapter 15: Problem 34
Find the general solution of the equation $$ x \frac{d^{3} y}{d x^{3}}+2 \frac{d^{2} y}{d x^{2}}=A x. $$
Short Answer
Expert verified
The general solution is \( y = -C_1 \text{ln}|x| + C_2 x + C_3 + \frac{A}{6} x^2 \).
Step by step solution
01
- Rewrite the Equation
Rewrite the given differential equation in a more convenient form: \[ x y''' + 2 y'' = Ax \]
02
- Solve the Homogeneous Equation
Consider the homogeneous part of the equation: \[ x y''' + 2 y'' = 0 \] This can be solved by making the substitution \( y'' = u \). Then we get: \[ x u' + 2u = 0 \]
03
- Solve for \( u \)
The homogeneous equation for \( u \) is: \[ x \frac{du}{dx} + 2u = 0 \] This can be solved by separation of variables: \[ \frac{1}{u} \frac{du}{dx} = -\frac{2}{x} \]Integrate both sides to find: \[ \text{ln} |u| = -2 \text{ln}|x| + C \]Or equivalently, \( u = \frac{C_1}{x^2} \). Since \( u = y'' \), we have: \[ y'' = \frac{C_1}{x^2} \]
04
- Integrate to Find \( y \)
Integrate \( y'' = \frac{C_1}{x^2} \) twice to find \( y \): \[ y' = -\frac{C_1}{x} + C_2 \] \[ y = -C_1 \text{ln}|x| + C_2 x + C_3 \]
05
- Solve the Non-Homogeneous Equation
The particular solution for the non-homogeneous equation \( x y''' + 2 y'' = Ax \) can be tried with a polynomial form. Assume a particular solution of the form \( y_p = B x^2 \).Substitute in to find: \[ x (2B) + 2 (2B) = Ax \] \[ 2B + 4B = A \] \[ 6B = A \Rightarrow B = \frac{A}{6} \]Thus, the particular solution is: \[ y_p = \frac{A}{6} x^2 \]
06
- Combine General Solution
Combine the general solution for the homogeneous equation and the particular solution: \[ y = -C_1 \text{ln}|x| + C_2 x + C_3 + \frac{A}{6} x^2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Homogeneous Equation
A homogeneous differential equation is one where every term depends on the unknown function and its derivatives.
In simpler terms, all the terms contain the dependent variable (usually y) or its derivatives.
Consider our example from the exercise. The homogeneous part of the equation is:
\[ x y''' + 2 y'' = 0 \]
This part is 'homogeneous' because there is no free term solely dependent on x.
To solve it, we first made a substitution: \( y'' = u \). This turns our third-order differential equation into a simpler first-order form:
\[ x u' + 2 u = 0 \]
This approach of turning higher-order equations into lower-order ones is common in solving homogeneous equations.
With this, we can now proceed to solve for \( u \) and eventually find \( y \).
In simpler terms, all the terms contain the dependent variable (usually y) or its derivatives.
Consider our example from the exercise. The homogeneous part of the equation is:
\[ x y''' + 2 y'' = 0 \]
This part is 'homogeneous' because there is no free term solely dependent on x.
To solve it, we first made a substitution: \( y'' = u \). This turns our third-order differential equation into a simpler first-order form:
\[ x u' + 2 u = 0 \]
This approach of turning higher-order equations into lower-order ones is common in solving homogeneous equations.
With this, we can now proceed to solve for \( u \) and eventually find \( y \).
Non-Homogeneous Equation
In a non-homogeneous differential equation, there is an extra term that is independent of the dependent variable.
This term is often called a 'forcing function' as it 'forces' the solution to take a particular form.
Look at our original equation again:
\[ x y''' + 2 y'' = Ax \]
The term \( Ax \) makes it non-homogeneous.
To solve such equations, we usually find the solution to the corresponding homogeneous equation first. Then we find a particular solution to the non-homogeneous equation.
These two solutions together give us the complete general solution.
In our particular case, the particular solution was assumed as \( y_p = B x^2 \). After determining \( B \), we arrive at the solution for the non-homogeneous part and combine it with our homogeneous solution.
This term is often called a 'forcing function' as it 'forces' the solution to take a particular form.
Look at our original equation again:
\[ x y''' + 2 y'' = Ax \]
The term \( Ax \) makes it non-homogeneous.
To solve such equations, we usually find the solution to the corresponding homogeneous equation first. Then we find a particular solution to the non-homogeneous equation.
These two solutions together give us the complete general solution.
In our particular case, the particular solution was assumed as \( y_p = B x^2 \). After determining \( B \), we arrive at the solution for the non-homogeneous part and combine it with our homogeneous solution.
Separation of Variables
Separation of variables is a method used to solve differential equations by separating the variables into two sides of the equation, each containing only one variable.
Let's use the example we started with:
We had to solve the homogeneous equation:
\[ x \frac{du}{dx} + 2 u = 0 \]
To solve this, we separated the variables involving x and u:
\[ \frac{1}{u} \frac{du}{dx} = -\frac{2}{x} \]
Then we integrated both sides:
\[ \text{ln} |u| = -2 \text{ln} |x| + C \]
This is how we got to the solution \( u = \frac{C_1}{x^2} \).
Separation of variables made solving this part of the differential equation simpler.
It's a straightforward and powerful technique for dealing with first-order differential equations.
Let's use the example we started with:
We had to solve the homogeneous equation:
\[ x \frac{du}{dx} + 2 u = 0 \]
To solve this, we separated the variables involving x and u:
\[ \frac{1}{u} \frac{du}{dx} = -\frac{2}{x} \]
Then we integrated both sides:
\[ \text{ln} |u| = -2 \text{ln} |x| + C \]
This is how we got to the solution \( u = \frac{C_1}{x^2} \).
Separation of variables made solving this part of the differential equation simpler.
It's a straightforward and powerful technique for dealing with first-order differential equations.
Particular Solution
A particular solution is a specific solution to a non-homogeneous differential equation, representing one instance that meets the equation.
We find this by assuming a form for the solution and determining any unknown constants.
In our example, we had the non-homogeneous differential equation:
\[ x y''' + 2 y'' = Ax \]
We assumed a solution of the form \( y_p = B x^2 \).
Substituting this into the equation and solving for B, we found:
\[ B = \frac{A}{6} \]
Hence, our particular solution is:
\[ y_p = \frac{A}{6} x^2 \]
Combining this with the general solution of the homogeneous part gives us the complete solution to our original differential equation.
Finding a particular solution is crucial in solving non-homogeneous differential equations, as it addresses the 'forcing function' making the equation non-homogeneous.
We find this by assuming a form for the solution and determining any unknown constants.
In our example, we had the non-homogeneous differential equation:
\[ x y''' + 2 y'' = Ax \]
We assumed a solution of the form \( y_p = B x^2 \).
Substituting this into the equation and solving for B, we found:
\[ B = \frac{A}{6} \]
Hence, our particular solution is:
\[ y_p = \frac{A}{6} x^2 \]
Combining this with the general solution of the homogeneous part gives us the complete solution to our original differential equation.
Finding a particular solution is crucial in solving non-homogeneous differential equations, as it addresses the 'forcing function' making the equation non-homogeneous.
