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Verify that \(y=e^{x}\) is a solution of the equation $$ (x-1) y^{\prime \prime}-x y^{\prime}+y=0 $$ Use this fact to find the general solution of $$ (x-1) y^{\prime \prime}-x y^{\prime}+y=1 $$.

Short Answer

Expert verified
The general solution is \( y = C_1 e^x + 1 \).

Step by step solution

01

Differentiate y

We are given \( y = e^x \). To verify it as a solution, we need to compute its derivatives. Start by calculating the first derivative: \( y' = \frac{d}{dx}e^x = e^x \).
02

Differentiate again

Now, find the second derivative: \( y'' = \frac{d}{dx}y' = \frac{d}{dx}e^x = e^x \).
03

Substitute into the differential equation

Substitute \( y, y', \) and \( y'' \) into the equation: \[ (x-1)y'' - xy' + y = (x-1)e^x - xe^x + e^x \].
04

Simplify the equation

Simplify the expression: \[ (x-1)e^x - xe^x + e^x = xe^x - e^x - xe^x + e^x = 0 \]. The equation simplifies correctly, verifying \( y = e^x \) is indeed a solution of the homogeneous equation \((x-1)y'' - xy' + y = 0\).
05

Solve the non-homogeneous equation

To solve \((x-1)y'' - xy' + y = 1\), first find a particular solution using the method of undetermined coefficients. Assume a constant solution \( y_p = C \), substituting gives \((x-1)(0) - x(0) + C = 1\), leading to \( C = 1 \). Thus, \( y_p = 1 \).
06

Write the general solution

The general solution of a non-homogeneous linear equation is given by the sum of the homogeneous solution and a particular solution. Therefore, the general solution is: \[ y = C_1 e^x + 1 \], where \( C_1 \) is an arbitrary constant.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Equation
In differential equations, a *homogeneous equation* is one where the sum of terms is set equal to zero. It often takes the form of a linear differential equation where all functions depend only on the independent variable and its derivatives. Homogeneous equations are significant because they form the foundation for solving more complex equations.
A core principle is the principle of *superposition*, which states that any linear combination of solutions to a homogeneous equation is also a solution. For example, if two functions, say \(y_1\) and \(y_2\), are solutions to a linear homogeneous equation, any function of the form \(C_1y_1 + C_2y_2\) will also be a solution, where \(C_1\) and \(C_2\) are constants.
To find specific solutions, you often start by assuming a certain form of the solution, like \(e^{rx}\) for linear differential equations with constant coefficients. The task then becomes finding the values of \(r\) that satisfy the equation.
Non-Homogeneous Equation
A *non-homogeneous equation* differs from its homogeneous counterpart because its right-hand side is non-zero. This means it has an additional term which often represents external forces or influences. These equations have the general form of:
\[Ly = f(x)\]
where \(L\) is a linear differential operator, \(y\) is the function of interest, and \(f(x)\) is a non-zero function.
Solving non-homogeneous equations requires finding both the homogeneous solution (complementary solution) as well as a particular solution. This combination provides the general solution. The key is first solving the associated homogeneous equation, then using methods like *undetermined coefficients* to find a particular solution that specifically addresses the non-homogeneity.
Undetermined Coefficients
When tackling a non-homogeneous differential equation, the *method of undetermined coefficients* is a strategic approach to find a specific solution. This technique works particularly well for equations where the non-homogeneous term, \(f(x)\), is a simple polynomial, exponential, or trigonometric function.
The essence of the method lies in making an educated guess about the form of the particular solution \(y_p\). The coefficients in this assumed form are termed as 'undetermined' because you solve for them later. For example, if \(f(x)\) is \(e^x\), you'd assume \(y_p = Ce^x\) where \(C\) is a constant. You then substitute \(y_p\) into the differential equation and solve for these coefficients, adjusting your initial guess if necessary.
General Solution
The *general solution* of a differential equation is a complete representation of all possible solutions. It is composed of two parts: the complementary or homogeneous solution and a particular solution. The complementary solution solves the associated homogeneous equation, while the particular solution addresses the non-homogeneous part.
In practice, finding the general solution involves:
  • Determining the *complementary solution* using methods like finding the roots of characteristic equations or via eigenfunctions.
  • Finding a *particular solution* using suitable tools like undetermined coefficients or variation of parameters.
The general solution is thereby formed as:
\[y = y_c + y_p\]
which involves taking a linear combination of the complementary and the particular solutions. This solution captures all individual solutions that satisfy the original differential equation.

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