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 the solution to the initial value problem.
$y'' + y' - 12 y = e^{t} + e^{2 t} - 1; y (0) = 1, y' (0) = 3$

The solution to the initial value problem is:

$y = \frac{7}{6} e^{3 t} + \frac{1}{60} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12}$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

$y'' + y' - 12 y = e^{t} + e^{2 t} - 1 . ..... (1)$

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

$y'' + y' - 12 y = 0$

The auxiliary equation for the above equation,

$\begin{matrix} m^{2} + m - 12 = 0 \\ m^{2} + 4 m - 3 m - 12 = 0 \\ m (m + 4) - 3 (m + 4) = 0 \\ (m - 3) (m + 4) = 0 \end{matrix}$

Step 2: Find the complementary solution of the given equation.

The root of an auxiliary equation is,

$m_{1} = 3, m_{2} = - 4$

The complementary solution of the given equation is,

$y_{c} = c_{1} e^{3 t} + c_{2} e^{- 4 t}$

Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

$y_{p} (t) = {Ae}^{t} + {Be}^{2 t} + C . .... (2)$

Now find the first and second derivatives of the above equation,

$\begin{matrix} y_{p}' (t) = {Ae}^{t} + 2 {Be}^{2 t} \\ y_{p}'' (t) = {Ae}^{t} + 4 {Be}^{2 t} \end{matrix}$

Substitute the value of $y_{p} (t), y_{p}' (t)$ and $y_{p}'' (t)$ the equation (1),

$\begin{matrix} y'' + y' - 12 y = e^{t} + e^{2 t} - 1 \\ {Ae}^{t} + 4 {Be}^{2 t} + {Ae}^{t} + 2 {Be}^{2 t} - 12 [{Ae}^{t} + {Be}^{2 t} + C] = e^{t} + e^{2 t} - 1 \\ - 10 {Ae}^{t} - 6 {Be}^{2 t} - 12 C = e^{t} + e^{2 t} - 1 \end{matrix}$

Comparing all coefficients of the above equation,

role="math" localid="1655098651285" $\begin{matrix} - 10 A = 1 \Rightarrow A = \frac{- 1}{10} \\ - 6 B = 1 \Rightarrow B = \frac{- 1}{6} \\ - 12 C = - 1 \Rightarrow C = \frac{1}{12} \end{matrix}$

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

$\begin{matrix} y_{p} (t) = {Ae}^{t} + {Be}^{2 t} + C \\ y_{p} (t) = - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12} \end{matrix}$

Step 4: Find the general solution and use the given initial condition.

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (t) + y_{p} (t) \\ y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12} . .... (3) \end{matrix}$

Given the initial condition,

$y (0) = 1, y' (0) = 3$

Substitute the value of y = 1 and t = 0 in the equation (3),

role="math" localid="1655099043521" $\begin{matrix} y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12} \\ 1 = c_{1} e^{3 (0)} + c_{2} e^{- 4 (0)} - \frac{1}{10} e^{0} - \frac{1}{6} e^{2 (0)} + \frac{1}{12} \\ 1 = c_{1} + c_{2} - \frac{1}{10} - \frac{1}{6} + \frac{1}{12} \\ c_{1} + c_{2} = \frac{71}{60} . ..... (4) \end{matrix}$

Now find the derivative of the equation (3),

role="math" localid="1655099101150" $y' = 3 c_{1} e^{3 t} - 4 c_{2} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{3} e^{2 t}$

Substitute the value of y’ = 3 and t = 0 in the above equation,

role="math" localid="1655099214255" $\begin{matrix} y' = 3 c_{1} e^{3 t} - 4 c_{2} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{3} e^{2 t} \\ 3 = 3 c_{1} e^{3 (0)} - 4 c_{2} e^{- 4 (0)} - \frac{1}{10} e^{0} - \frac{1}{3} e^{2 (0)} \\ 3 = 3 c_{1} - 4 c_{2} - \frac{1}{10} - \frac{1}{3} \\ 3 c_{1} - 4 c_{2} = 3 + \frac{13}{30} \\ 3 c_{1} - 4 c_{2} = \frac{103}{30} . ..... (5) \end{matrix}$

Solve the equation (4) and (5),

role="math" localid="1655099327421" $\begin{matrix} 4 (c_{1} + c_{2}) = \frac{71}{60} \times 4 \\ 4 c_{1} + 4 c_{2} = \frac{71}{15} \\ 3 c_{1} - 4 c_{2} = \frac{103}{30} \\ c_{1} = \frac{35}{30} \\ c_{1} = \frac{7}{6} \end{matrix}$

Substitute the value of $c_{1}$ in the equation (4),

role="math" localid="1655099442348" $\begin{matrix} c_{1} + c_{2} = \frac{71}{60} \\ \frac{7}{6} + c_{2} = \frac{71}{60} \\ c_{2} = \frac{71}{60} - \frac{7}{6} \\ c_{2} = \frac{1}{60} \end{matrix}$

Substitute the value of $c_{1}$ and $c_{2}$ in the equation (3),

role="math" localid="1655099619782" $\begin{matrix} y = c_{1} e^{3 t} + c_{2} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12} \\ y = \frac{7}{6} e^{3 t} + \frac{1}{60} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12} \end{matrix}$

Thus, the solution to the initial value problem is:

role="math" localid="1655099676357" $y = \frac{7}{6} e^{3 t} + \frac{1}{60} e^{- 4 t} - \frac{1}{10} e^{t} - \frac{1}{6} e^{2 t} + \frac{1}{12}$

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Find the solution to the initial value problem.
$y'' + y' - 12 y = e^{t} + e^{2 t} - 1; y (0) = 1, y' (0) = 3$

Step by Step Solution

Step 1: Write the auxiliary equation of the given differential equation.

Step 2: Find the complementary solution of the given equation.

Step 3: Now find the particular solution to find a general solution for the equation.

Step 4: Find the general solution and use the given initial condition.

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Find the solution to the initial value problem.y''+y'-12y=et+e2t-1; y(0)=1, y'(0)=3

Step by Step Solution

Step 1: Write the auxiliary equation of the given differential equation.

Step 2: Find the complementary solution of the given equation.

Step 3: Now find the particular solution to find a general solution for the equation.

Step 4: Find the general solution and use the given initial condition.

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Find the solution to the initial value problem.
$y'' + y' - 12 y = e^{t} + e^{2 t} - 1; y (0) = 1, y' (0) = 3$