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. $y'' (x) + y (x) = 2^{x}$

Thus, the particular solution is $y_{p} = 2^{x} {({[\ln (2)]}^{2} + 1)}^{- 1}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Use logarithms properties for simplification of the given differential equation.

The given differential equation is:

$y'' (x) + y (x) = 2^{x}$

Simplify the above equation by using logarithms properties,

$\begin{matrix} y'' (x) + y (x) = e^{\ln (2^{x})} \\ y'' (x) + y (x) = e^{x [\ln (2)]} \\ y'' (x) + y (x) = e^{[\ln (2)] x} \end{matrix}$

Step 2: Firstly, write the auxiliary equation of the above differential equation.

The auxiliary equation for the above equation:

$m^{2} + 1 = 0$

Step 3: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{matrix} m^{2} + 1 = 0 \\ m^{2} = - 1 \\ m = \sqrt{- 1} \\ m = \pm i \end{matrix}$

The roots of the auxiliary equation are:

$m_{1} = i & m_{2} = - i$

The complementary solution of the given equation is:

$y_{c} = c_{1} \cos (x) + c_{2} \sin (x)$

Step 4: Final conclusion, find a particular solution to the differential equation.

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

$y_{p} = {Ae}^{[\ln (2)] x} \dots (2)$

Now find the derivative of the above equation,

$\begin{matrix} y_{p}' = Aln (2) e^{[\ln (2)] x} \\ y_{p}'' = A {[\ln (2)]}^{2} e^{[\ln (2)] x} \end{matrix}$

From the equation (1),

$\begin{matrix} y_{p}'' + y_{p} = e^{[\ln (2)] x} \\ A {[\ln (2)]}^{2} e^{[\ln (2)] x} + {Ae}^{[\ln (2)] x} = e^{[\ln (2)] x} \\ {Ae}^{[\ln (2)] x} [{[\ln (2)]}^{2} + 1] = e^{[\ln (2)] x} \\ A [{[\ln (2)]}^{2} + 1] = 1 \\ A = \frac{}{{[\ln (2)]}^{2} + 1} \end{matrix}$

Substitute the value of A in the equation (2), we get:

$y_{p} = 2^{x} {({[\ln (2)]}^{2} + 1)}^{- 1}$

Therefore, the particular solution of equation (1),

$\begin{matrix} y_{p} = (\frac{1}{{[\ln (2)]}^{2} + 1}) e^{[\ln (2)] x} \\ y_{p} = \frac{2^{x}}{{[\ln (2)]}^{2} + 1} \\ y_{p} = 2^{x} {({[\ln (2)]}^{2} + 1)}^{- 1} \end{matrix}$

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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. $y'' (x) + y (x) = 2^{x}$

Step by Step Solution

Step 1: Use logarithms properties for simplification of the given differential equation.

Step 2: Firstly, write the auxiliary equation of the above differential equation.

Step 3: Now find the roots of the auxiliary equation.

Step 4: Final conclusion, find a particular solution to the differential equation.

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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. y''(x)+y(x)=2x

Step by Step Solution

Step 1: Use logarithms properties for simplification of the given differential equation.

Step 2: Firstly, write the auxiliary equation of the above differential equation.

Step 3: Now find the roots of the auxiliary equation.

Step 4: Final conclusion, find a particular solution to the differential equation.

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Find a particular solution to the differential equation. $y'' (x) + y (x) = 2^{x}$