Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

Find a particular solution to the differential equation.
$\frac{d^{2} y}{{dx}^{2}} - 5 \frac{dy}{dx} + 6 y = {xe}^{x}$

The particular solution of the differential equation is $y_{p} = (\frac{1}{2} x + \frac{3}{4}) e^{x} .$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Firstly, write the auxiliary equation of the above differential equation:

We have,

$\frac{d^{2} y}{{dx}^{2}} - 5 \frac{dy}{dx} + 6 y = {xe}^{x} ...... (1)$

Write the homogeneous differential equation of the equation (1),

$\frac{d^{2} y}{{dx}^{2}} - 5 \frac{dy}{dx} + 6 y = 0$

The auxiliary equation for the above equation,

$m^{2} - 5 m + 6 = 0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{matrix} m^{2} - 5 m + 6 = 0 \\ m^{2} - 3 m - 2 m + 6 = 0 \\ m (m - 3) - 2 (m - 3) = 0 \\ (m - 3) (m - 2) = 0 \end{matrix}$

The roots of the auxiliary equation are,

$m_{1} = 2, & m_{2} = 3$

The complementary solution of the given equation is,

$y_{c} = c_{1} e^{2 x} + c_{2} e^{3 x}$

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

$y_{p} = (Ax + B) e^{x} . ..... (2)$

Now find the derivative of the above equation,

$\begin{matrix} y_{p}' = (A) e^{x} + (Ax + B) e^{x} \\ y_{p}'' = (A) e^{x} + (A) e^{x} + (Ax + B) e^{x} \\ = 2 (A) e^{x} + (Ax + B) e^{x} \end{matrix}$

From the equation (1), Substitute the value of $y_{p}'', y_{p}'$ and $y_{p}$ in the equation (1),

$\begin{matrix} \frac{d^{2} y}{{dx}^{2}} - 5 \frac{}{} + 6 y = {xe}^{x} \\ 2 (A) e^{x} + (Ax + B) e^{x} - 5 ((A) e^{x} + (Ax + B) e^{x}) + 6 ((Ax + B) e^{x}) = {xe}^{x} \\ - 3 (A) e^{x} + 2 (Ax + B) e^{x} = {xe}^{x} \\ 2 {Axe}^{x} + (- 3 A + 2 B) e^{x} = {xe}^{x} \end{matrix}$

Step 4: Final conclusion

Comparing all coefficients of the above equation;

$\begin{matrix} 2 A = 1 \\ A = \frac{1}{2} \\ - 3 A + 2 B = 0 . ..... (3) \end{matrix}$

Substitute the value of Ain the equation (3),

$\begin{matrix} - 3 (\frac{1}{2}) + 2 B = 0 \\ B = \frac{3}{4} \end{matrix}$

Substitute the value of A and B in the equation (2),

$y_{p} = (\frac{1}{2} x + \frac{3}{4}) e^{x}$

Therefore, the particular solution of equation (1);

$y_{p} = (\frac{1}{2} x + \frac{3}{4}) e^{x}$

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find a particular solution to the differential equation.
$\frac{d^{2} y}{{dx}^{2}} - 5 \frac{dy}{dx} + 6 y = {xe}^{x}$

Step by Step Solution

Step 1: Firstly, write the auxiliary equation of the above differential equation:

Step 2: Now find the roots of the auxiliary equation

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

Step 4: Final conclusion

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find a particular solution to the differential equation.d2ydx2-5dydx+6y=xex

Step by Step Solution

Step 1: Firstly, write the auxiliary equation of the above differential equation:

Step 2: Now find the roots of the auxiliary equation

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

Step 4: Final conclusion

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Find a particular solution to the differential equation.
$\frac{d^{2} y}{{dx}^{2}} - 5 \frac{dy}{dx} + 6 y = {xe}^{x}$