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Mobile App Chapter 12: Problem 22
For each of the following equations, one solution \(u\) is given. Find the other solution by assuming a solution of the form \(y=u v\). \(x^{2}(2-x) y^{\prime \prime}+2 x y^{\prime}-2 y=0 ; u=x\)
Short Answer
Expert verified
The second solution is \( y = x(-\text{ln}|2-x| + K) \).
Step by step solution
01
Introduce the form of the solution
Given that one solution is assumed, take the general form of the solution as: \[ y = uv \] where \( u = x \) and \( v \) needs to be determined.
02
Compute the first and second derivatives of y
First, compute the first derivative of \( y \): \[ y' = (uv)' = u'v + uv' = x v' + v \] Next, compute the second derivative of \( y \): \[ y'' = (x v' + v)' = x v'' + v' + v' = x v'' + 2v' \]
03
Substitute derivatives into the original equation
Substitute \( y = x v \), \( y' = x v' + v \), and \( y'' = x v'' + 2v' \) into the differential equation: \[ x^{2}(2-x)(x v'' + 2v') + 2x(x v' + v) - 2(x v) = 0 \]
04
Simplify the equation
Distribute and combine like terms: \[ x^3(2-x)v'' + 2x^2(2-x)v' + 2x^2v' + 2xv - 2xv = 0 \] Simplify further as: \[ x^3(2-x)v'' + 4x^2v' = 0 \]
05
Factor and solve the simplified equation
Factor out common terms: \[ x^2 v' (v''(2-x) + 4) = 0 \] This implies two cases: \[ x^2 eq 0 \rightarrow v''(2-x) + 4 = 0 \] Solve for \( v \): \[ v''(2-x) + 4 = 0 \implies v'' = -\frac{4}{2-x} \]
06
Integrate to find v
Integrate \( v'' = -\frac{4}{2-x} \): \[ v' = \frac{4 \text{ ln}|2-x|}{-1} + C_1 \] \[ v' = -4\text{ln}|2-x| + C_1 \] Integrate again: \[ v = -4 \text{Li}(x-2) + C_1 x + C_2 \]
07
Form the complete solution
Apply initial conditions or constraints if any: After simplifying, the second solution can be written in a general form, hence: \[ y = x v = x(-\text{ln}|2-x| + K) \] where \( K \) is a constant.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
second-order differential equation
A second-order differential equation involves derivatives of a function up to the second order. These types of equations are crucial in modeling diverse physical phenomena like vibrations, electrical circuits, and mechanical systems.
In a second-order differential equation, the highest derivative is the second derivative. For example:
\[ x^2 y'' + 2x y' - 2 y = 0 \]
This particular equation can model the behavior of different systems, including mechanical oscillations or electrical circuits. Understanding these equations is vital for predicting how a system responds to various influences.
In a second-order differential equation, the highest derivative is the second derivative. For example:
\[ x^2 y'' + 2x y' - 2 y = 0 \]
This particular equation can model the behavior of different systems, including mechanical oscillations or electrical circuits. Understanding these equations is vital for predicting how a system responds to various influences.
solution by substitution
Solution by substitution is a method used to simplify and solve differential equations. This technique involves replacing the original function with a product of simpler functions, which often makes the equation easier to handle.
In our exercise, we assume the solution in the form \( y = u v \). With one known solution \( u = x \), we substitute and then determine the unknown \( v \).
By computing the first and second derivatives and substituting them into the original equation, we transform a complex differential equation into a more manageable form. This approach breaks down the problem and can reveal relationships between different parts of the equation.
In our exercise, we assume the solution in the form \( y = u v \). With one known solution \( u = x \), we substitute and then determine the unknown \( v \).
By computing the first and second derivatives and substituting them into the original equation, we transform a complex differential equation into a more manageable form. This approach breaks down the problem and can reveal relationships between different parts of the equation.
integral calculus
Integral calculus plays a vital role in solving differential equations, especially when finding particular solutions.
For instance, after substituting and differentiating, we often need to integrate to find functions like \( v \).
In the given problem, after determining \( v'' \), we integrate to find \( v' \) and then integrate again to find \( v \).
One of the integrals we encounter is:
\[ v' = -4 \ln|2-x| + C_1 \]
By integrating \( v' \), we obtain:
\[ v = -4 \text{Li}(x-2) + C_1 x + C_2 \]
This step-by-step integration process is fundamental in transitioning from the differential form to an explicit solution.
For instance, after substituting and differentiating, we often need to integrate to find functions like \( v \).
In the given problem, after determining \( v'' \), we integrate to find \( v' \) and then integrate again to find \( v \).
One of the integrals we encounter is:
\[ v' = -4 \ln|2-x| + C_1 \]
By integrating \( v' \), we obtain:
\[ v = -4 \text{Li}(x-2) + C_1 x + C_2 \]
This step-by-step integration process is fundamental in transitioning from the differential form to an explicit solution.
homogeneous differential equation
A homogeneous differential equation is one in which every term is a multiple of the function or its derivatives. All homogeneous equations have solutions that can be expressed in terms of constants and exponential functions.
The given equation:
\[ x^2(2-x) y'' + 2x y' - 2y = 0 \]
is homogeneous because each term involves the unknown function \( y \) or its derivatives combined with variables.
Homogeneous equations often have simpler solving methods, and their solutions can reveal insights into the system modeled by the equation. Finding these solutions typically involves assuming a form for the solution and substituting back into the equation, then solving the resulting algebraic equations.
The given equation:
\[ x^2(2-x) y'' + 2x y' - 2y = 0 \]
is homogeneous because each term involves the unknown function \( y \) or its derivatives combined with variables.
Homogeneous equations often have simpler solving methods, and their solutions can reveal insights into the system modeled by the equation. Finding these solutions typically involves assuming a form for the solution and substituting back into the equation, then solving the resulting algebraic equations.
