Q27E Find the solution to the initial... [FREE SOLUTION]

Chapter 4: Q27E (page 186)

Find the solution to the initial value problem.
$y'' (x) - y' (x) - 2 y (x) = cosx - \sin 2 x; 鈥夆赌夆赌夆赌夆赌� y (0) = \frac{- 7}{20}, 鈥夆赌夆赌夆赌� y' (0) = \frac{1}{5}$

Short Answer

Expert verified

The initial solution to the differential equation is: $y = - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x)$

Step by step solution

Write the auxiliary equation of the given differential equation.

The differential equation is,

$y'' (x) - y' (x) - 2 y (x) = cosx - \sin 2 x 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€� . ..... (1)$

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

$y'' (x) - y' (x) - 2 y (x) = 0$

The auxiliary equation for the above equation,

$m^{2} - m - 2 = 0$

Find the complementary solution of the given equation.

Solve the above equation,

$\begin{matrix} m^{2} - m - 2 = 0 \\ m^{2} - 2 m + m - 2 = 0 \\ m (m - 2) + 1 (m - 2) = 0 \\ (m - 2) (m + 1) = 0 \end{matrix}$

The root of an auxiliary equation is,

$m_{1} = 2, 鈥夆赌夆赌� m_{2} = - 1$

The complementary solution of the given equation is,

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

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

Assume, the particular solution of equation (1),

$y_{p} (x) = Acosx + Bsin (2 x) + Csinx + Dcos (2 x) 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌� . ..... (2)$

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

$\begin{matrix} y_{p}' (x) = - Asinx + 2 Bcos (2 x) + Ccosx - 2 Dsin (2 x) \\ y_{p}'' (x) = - Acosx - 4 Bsin (2 x) - Csinx - 4 Dcos (2 x) \end{matrix}$

Substitute the value of $y (x), 鈥夆赌� y' (x)$ and $y_{p}'' (t)$ the equation (1),

$\begin{array}{l} 鈬� y'' (x) - y' (x) - 2 y (x) = cosx - \sin 2 x \\ 鈬� - Acosx - 4 Bsin (2 x) - Csinx - 4 Dcos (2 x) - [- Asinx + 2 Bcos (2 x) + Ccosx - 2 Dsin (2 x)] \\ 鈥夆赌夆赌夆赌夆赌夆€� - 2 [Acosx + Bsin (2 x) + Csinx + Dcos (2 x)] = cosx - \sin 2 x \\ 鈬� (- 3 A - C) cosx + (- 3 C + A) sinx + (- 6 D - 2 B) \cos (2 x) + (- 6 B + 2 D) \sin (2 x) = cosx - \sin 2 x \end{array}$

Comparing all coefficients of the above equation,

$\begin{matrix} - 3 A - C = 1 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌� . ..... (3) \\ - 6 B + 2 D = - 1 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌� . ..... (4) \\ - 3 C + A = 0 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€� . ..... (5) \\ - 6 D - 2 B = 0 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌� . ..... (6) \end{matrix}$

Solve the equation (3) and (5),

role="math" localid="1655096234005" $\begin{matrix} 3 (- 3 A - C) = 1 脳 3 \\ - 3 C - 9 A = 3 \\ 鈥� - 3 C + A = 0 \\ 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌� A = \frac{- 3}{10} \end{matrix}$

Substitute the value of A in the equation (3),

role="math" localid="1655096019737" $\begin{matrix} - 3 A - C = 1 \\ - 3 (\frac{- 3}{10}) - C = 1 \\ c = \frac{9}{10} - 1 \\ c = \frac{- 1}{10} \end{matrix}$

Solve the equation (4) and (6),

role="math" localid="1655096203374" $3 (- 6 B + 2 D) = 鈭� 1 脳 3 \begin{matrix} 6 D - 18 B = - 3 \\ - 6 D - 2 B = 0 \\ B = \frac{3}{20} \end{matrix}$

Substitute the value of B in the equation (4),

role="math" localid="1655096279255" $\begin{matrix} - 6 B + 2 D = - 1 \\ - 6 (\frac{3}{20}) + 2 D = - 1 \\ 2 D = - 1 + \frac{9}{10} \\ D = \frac{- 1}{20} \end{matrix}$

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

$\begin{matrix} y_{p} (x) = Acosx + Bsin (2 x) + Csinx + Dcos (2 x) \\ y_{p} (x) = - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x) \end{matrix}$

Find the general solution and use the given initial condition.

Therefore, the general solution is,

$\begin{matrix} y = y_{c} (x) + y_{p} (x) \\ y = c_{1} e^{2 x} + c_{2} e^{- x} - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x) 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌� . .... (7) \end{matrix}$

Given the initial condition,

$y (0) = \frac{- 7}{20}, 鈥夆赌夆赌夆赌� y' (0) = \frac{1}{5}$

Substitute the value of $y = \frac{- 7}{20}$ and x = 0 in the equation (7),

$\begin{matrix} \frac{- 7}{20} = c_{1} e^{2 (0)} + c_{2} e^{- 0} - \frac{3}{10} \cos (0) + \frac{3}{20} \sin (0) - \frac{1}{10} \sin (0) - \frac{1}{20} \cos (0) \\ \frac{- 7}{2} = c_{1} + c_{2} - \frac{3}{10} - \frac{1}{20} \\ c_{1} + c_{2} = - \frac{7}{20} + \frac{3}{10} + \frac{1}{20} \\ c_{1} + c_{2} = 0 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€� . ..... (8) \end{matrix}$

Now find the derivative of the equation (7),

$y' = 2 c_{1} e^{2 x} - c_{2} e^{- x} + \frac{3}{10} sinx + \frac{3}{10} \cos (2 x) - \frac{1}{10} cosx + \frac{1}{10} \sin (2 x)$

Substitute the value of $y' = \frac{1}{5}$ and x = 0 in the above equation,

$\begin{matrix} \frac{1}{5} = 2 c_{1} e^{2 (0)} - c_{2} e^{- 0} + \frac{3}{10} \sin (0) + \frac{3}{10} \cos (0) - \frac{1}{10} \cos (0) + \frac{1}{10} \sin (0) \\ \frac{1}{5} = 2 c_{1} - c_{2} + \frac{3}{10} - \frac{1}{10} \\ 2 c_{1} - c_{2} = \frac{1}{5} - \frac{3}{10} + \frac{1}{10} \\ 2 c_{1} - c_{2} = 0 鈥夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌夆赌夆赌夆€夆赌夆赌� . ..... (9) \end{matrix}$

Solve the equation (8) and (9),

$\begin{matrix} c_{1} + c_{2} = 0 \\ 2 c_{1} - c_{2} = 0 \\ c_{1} = 0 \end{matrix}$

Substitute the value of $C_{1} = 0$ in the equation (8),

$\begin{matrix} c_{1} + c_{2} = 0 \\ c_{2} = 0 \end{matrix}$

Substitute the value of $c_{1} = 0$ and $c_{2} = 0$ in the equation (7),

role="math" localid="1655097899726" $\begin{matrix} y = (0) e^{2 x} + (0) e^{- x} - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x) \\ y = - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x) \end{matrix}$

Thus, the initial solution to the differential equation is:

$y = - \frac{3}{10} cosx + \frac{3}{20} \sin (2 x) - \frac{1}{10} sinx - \frac{1}{20} \cos (2 x)$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

91影视

Find the solution to the initial value problem.
$y'' (x) - y' (x) - 2 y (x) = cosx - \sin 2 x; 鈥夆赌夆赌夆赌夆赌� y (0) = \frac{- 7}{20}, 鈥夆赌夆赌夆赌� y' (0) = \frac{1}{5}$

Short Answer

Step by step solution

Write the auxiliary equation of the given differential equation.

Find the complementary solution of the given equation.

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

Find the general solution and use the given initial condition.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Probability and Statistics

Theoretical and Mathematical Physics

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

91影视

Find the solution to the initial value problem.y''(x)-y'(x)-2y(x)=cosx-sin2x;鈥夆赌夆赌夆赌夆赌�y(0)=-720,鈥夆赌夆赌夆赌�y'(0)=15

Short Answer

Step by step solution

Write the auxiliary equation of the given differential equation.

Find the complementary solution of the given equation.

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

Find the general solution and use the given initial condition.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Statistics

Probability and Statistics

Theoretical and Mathematical Physics

Discrete Mathematics

Logic and Functions

Study anywhere. Anytime. Across all devices.

Find the solution to the initial value problem.
$y'' (x) - y' (x) - 2 y (x) = cosx - \sin 2 x; 鈥夆赌夆赌夆赌夆赌� y (0) = \frac{- 7}{20}, 鈥夆赌夆赌夆赌� y' (0) = \frac{1}{5}$