Question

Solve the initial value problem, given that \(y_{1}\) satisfies the complementary equation. \((x+1)^{2} y^{\prime \prime}-2(x+1) y^{\prime}-\left(x^{2}+2 x-1\right) y=(x+1)^{3} e^{x}, \quad y(0)=1, \quad y^{\prime}(0)=-1 ;\) \(y_{1}=(x+1) e^{x}\)

Short answer

Question: Given that \(y_1=(x+1)e^x\) satisfies the complementary equation for the second-order linear non-homogeneous differential equation \((x+1)^2 y'' - 2(x+1) y' - (x^2+ 2x -1) y= (x+1)^3 e^x\), and the initial conditions \(y(0) = 1\) and \(y'(0)=-1\), determine the solution to the initial value problem using the method of variation of parameters. Answer: The solution to the initial value problem is given by: \(y(x) = C_{1}(x+1)e^x + u(x)(x+1)e^x\), where \(C_1\) is obtained from initial conditions and \(u(x)\) is obtained by integrating: \(u'(x) = \int \frac{(x+1)^3}{(x+1)^3 + x(-x^2 + x)} \, dx\) and then, \(u(x) = \int u'(x) \, dx\).

Step-by-step solution

Complementary solution

Because it is given that \(y_{1}=(x+1)e^{x}\) satisfies the complementary equation, we can write the complementary solution as: \(y_c = C_{1}(x+1)e^x\)

Variation of parameters

Now, we'll find the particular solution using the method of variation of parameters. First, we'll guess that the particular solution takes the form: \(y_p = u(x)(x+1)e^x\) Then, we need to find the first and second derivatives of \(y_p\): \(y_p' = u'(x)(x+1)e^x + u(x)e^x\) \(y_p'' = u''(x)(x+1)e^x + 2u'(x)e^x\)

Substitute into the differential equation

Now, we can substitute \(y_p\), \(y_p'\), and \(y_p''\) into the given differential equation: \((x+1)^{2}(u''(x)(x+1)e^x + 2u'(x)e^x) - 2(x+1)(u'(x)(x+1)e^x + u(x)e^x) - (x^{2}+2 x-1)(u(x)(x+1)e^x) = (x+1)^3 e^x\) After simplification, we obtain: \(u''(x)(x+1)^3e^x + 2u'(x)(x+1)^2e^x - 2u'(x)(x+1)^2e^x - 2u(x)(x+1)e^x + u(x)(x^2+2x-1)e^x = (x+1)^3 e^x\) Notice that some terms cancel out. Now, the equation becomes: \(u''(x)(x+1)^3e^x - u(x)(x+1)e^x + u(x)(x^2+2x-1)e^x = (x+1)^3 e^x\)

Simplify the equation

Now, we can collect the terms containing \(u(x)\) and \(u''(x)\): \(u''(x)(x+1)^3e^x + u(x)(-x^2 + x)e^x = (x+1)^3 e^x\) Now, we can divide by the common factor \(e^x\): \(u''(x)(x+1)^3 + u(x)(-x^2 + x) = (x+1)^3\)

Integrate to find \(u(x)\)

In order to find \(u(x)\), we need to integrate twice. Let's first rewrite this equation as: \(u''(x) = \frac{(x+1)^3}{(x+1)^3 + x(-x^2 + x)}\) We integrate once with respect to x to get \(u'(x)\): \(u'(x) = \int \frac{(x+1)^3}{(x+1)^3 + x(-x^2 + x)} \, dx\) We integrate again with respect to x to get \(u(x)\): \(u(x) = \int u'(x) \, dx\)

Find the general solution

Now, we have found \(u(x)\). We can use this to find the particular solution: \(y_p = u(x)(x+1)e^x\) Then, we can add the complementary solution and the particular solution to find the general solution: \(y(x) = y_c + y_p = C_{1}(x+1)e^x + u(x)(x+1)e^x\)

Apply the initial conditions

Finally, we will use the given initial conditions to find the constant \(C_1\). Apply the initial conditions to the general solution: \(y(0) = C_{1}(0+1)e^{0} + u(0)(0+1)e^{0} = 1\) \(y'(0) = -1\) Solve this system of equations to find the value of \(C_1\). Then, substitute the value of \(C_1\) back into the general solution to get the final answer.

Key concepts

Complementary Solution

The complementary solution of a differential equation is the solution to the associated homogeneous equation, which means setting the non-homogeneous part of the equation to zero. In simpler terms, we only consider the parts of the equation that involve the derivatives and the function itself. Here, the complementary solution was provided as part of the problem statement. It was given that a function, \( y_{1} = (x+1)e^{x} \), satisfies the complementary equation.
Therefore, the complementary solution represents the family of solutions that can be scaled by a constant, \( C_1 \). This is written as:

• \( y_c = C_1 (x+1) e^x \)

This solution forms an integral part of calculating the general solution of the differential equation, as it accounts for the behavior of the system without external inputs or forcing functions.

Variation of Parameters

Variation of parameters is a method used to find a particular solution to a non-homogeneous differential equation. This approach adjusts the constants in the complementary solution to be functions of the independent variable, allowing the solution to account for the non-homogeneous part. In this problem, we assumed the particular solution has the form:

• \( y_p = u(x)(x+1)e^x \)

The strategy involves allowing \( u(x) \) to vary such that the particular solution satisfies the non-homogeneous equation. First, we find the derivatives of \( y_p \) to substitute into the original equation. The method seeks to determine \( u(x) \) by simplifying the resulting equation, isolating the terms in \( u(x) \) and distinguishing between those that need integration. This allows us to systematically manage the complexity brought on by non-homogeneous terms.

Particular Solution

The particular solution addresses the specific conditions of the non-homogeneous differential equation, tailoring the solution to account for the forcing function. Once the general form of the solution \( y_p = u(x)(x+1)e^x \) is assumed, the next step is to determine \( u(x) \) through integration.
The process involves manipulating the differential equation so that terms involving \( u(x) \) can be integrated, thereby yielding expressions for \( u'(x) \) and subsequently \( u(x) \) itself. At the end of this integration step, \( y_p \) can be accurately formulated, equipped to nullify the non-homogeneous aspects of the equation and satisfy the original differential equation's particular challenges.

General Solution

The general solution of a differential equation combines both the complementary and the particular solutions, incorporating all possible solutions of the given differential equation. In essence, it is expressed as:

• \( y(x) = y_c + y_p \)
• \( y(x) = C_1 (x+1)e^x + u(x)(x+1)e^x \)

This combination ensures that any external influences represented by the non-homogeneous term are counterbalanced with the particular solution, while the complementary solution forms the basis of the solution set.\[ \text{General Solution} = C_1 (x+1)e^x + u(x)(x+1)e^x \]Initial conditions then guide the finalization of this solution by pinpointing the specific constants required – like \( C_1 \) – to ensure that the solution aligns precisely with specific conditions such as given function values or derivatives at particular points.