Please answer in detail. Thank you **Problem:**Find the general solution of the differential equation:\[y'' - 4y' + 4y = \frac{e^{2x}}{x}\]**Explanation:**This is a linear second-order non-homogeneous differential equation. The left-hand side is the homogeneous part, and the right-hand side is the non-homogeneous part. To find the general solution, follow these steps:1. **Solve the homogeneous equation:** - The homogeneous equation is $ y'' - 4y' + 4y = 0 $. - Find the characteristic equation: $ r^2 - 4r + 4 = 0 $. - Solve for $ r $ to find the roots. Since $ (r-2)^2 = 0 $, the roots are $ r = 2 $, a repeated root. - The complementary solution $ y_c $ is given by: \[ y_c = C_1 e^{2x} + C_2 x e^{2x} \]2. **Solve the non-homogeneous equation using appropriate methods:** - Since the right-hand side is $ \frac{e^{2x}}{x} $, use the method of variation of parameters or another appropriate method for solving non-homogeneous equations with a term involving $ x $.3. **Combine solutions:** - The general solution is a combination of the homogeneous solution and a particular solution $ y_p $: \[ y = y_c + y_p = C_1 e^{2x} + C_2 x e^{2x} + y_p \]**Note:** You may need to perform calculations to find the particular solution $ y_p $.

Answered: Find the general solution of: y"- 4y' +…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Please answer in detail. Thank you

**Problem:**

Find the general solution of the differential equation:

\[
y'' - 4y' + 4y = \frac{e^{2x}}{x}
\]

**Explanation:**

This is a linear second-order non-homogeneous differential equation. The left-hand side is the homogeneous part, and the right-hand side is the non-homogeneous part. To find the general solution, follow these steps:

1. **Solve the homogeneous equation:**
- The homogeneous equation is $ y'' - 4y' + 4y = 0 $.
- Find the characteristic equation: $ r^2 - 4r + 4 = 0 $.
- Solve for $ r $ to find the roots. Since $ (r-2)^2 = 0 $, the roots are $ r = 2 $, a repeated root.
- The complementary solution $ y_c $ is given by:
\[
y_c = C_1 e^{2x} + C_2 x e^{2x}
\]

2. **Solve the non-homogeneous equation using appropriate methods:**
- Since the right-hand side is $ \frac{e^{2x}}{x} $, use the method of variation of parameters or another appropriate method for solving non-homogeneous equations with a term involving $ x $.

3. **Combine solutions:**
- The general solution is a combination of the homogeneous solution and a particular solution $ y_p $:
\[
y = y_c + y_p = C_1 e^{2x} + C_2 x e^{2x} + y_p
\]

**Note:** You may need to perform calculations to find the particular solution $ y_p $.

Expert Solution

Step 1

Method of variation of parameters :

$consider the differential equation \frac{d^{2} y}{d x^{2}} + P (x) \frac{d y}{d x} + Q (x) y = R (x) . . . . . . . . . . . . . . . . . . . . . . . (1) Step 1 : Find two independent solution of the equation \frac{d^{2} y}{d x^{2}} + P (x) \frac{d y}{d x} + Q (x) y = 0 and denote them y_{1} and y_{2} . step 2 : Put C . F = {Ay}_{1} + {By}_{2}, where A and B are arbitrray constant . step 3 : Find W = |\begin{matrix} y_{1} & y_{2} \\ y_{1}' & y_{2}' \end{matrix}|, u = - \int \frac{y_{2} R}{W} dx + c_{1} and v = \int \frac{y_{1} R}{W} dx + c_{2} step 4 : Replace arbitrary constant A and B in C . F = {Ay}_{1} + {By}_{2} by functions u and v so that the complete solutions is y = {uy}_{1} + {vy}_{2} .$