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Expert-verified Found in: Page 186 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Find the solution to the initial value problem.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{-}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{-}}{\mathbf{2}}{\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}{\mathbf{cosx}}{\mathbf{-}}{\mathbf{sin}}{\mathbf{2}}{\mathbf{x}}{\mathbf{;}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}\frac{-7}{20}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}\frac{1}{5}$

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

See the step by step solution

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

The differential equation is,

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=\mathrm{cosx}-\mathrm{sin}2\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

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

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=0$

The auxiliary equation for the above equation,

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

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

Solve the above equation,

$\begin{array}{c}{\mathrm{m}}^{2}-\mathrm{m}-2=0\\ {\mathrm{m}}^{2}-2\mathrm{m}+\mathrm{m}-2=0\\ \mathrm{m}\left(\mathrm{m}-2\right)+1\left(\mathrm{m}-2\right)=0\\ \left(\mathrm{m}-2\right)\left(\mathrm{m}+1\right)=0\end{array}$

The root of an auxiliary equation is,

${\mathrm{m}}_{1}=2,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-1$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{x}}$

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

Assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=-\mathrm{Asinx}+2\mathrm{Bcos}\left(2\mathrm{x}\right)+\mathrm{Ccosx}-2\mathrm{Dsin}\left(2\mathrm{x}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=-\mathrm{Acosx}-4\mathrm{Bsin}\left(2\mathrm{x}\right)-\mathrm{Csinx}-4\mathrm{Dcos}\left(2\mathrm{x}\right)\end{array}$

Substitute the value of $\mathrm{y}\left(\mathrm{x}\right),\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\text{'}\left(\mathrm{x}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ the equation (1),

$\begin{array}{l}⇒\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-\mathrm{y}\text{'}\left(\mathrm{x}\right)-2\mathrm{y}\left(\mathrm{x}\right)=\mathrm{cosx}-\mathrm{sin}2\mathrm{x}\\ ⇒-\mathrm{Acosx}-4\mathrm{Bsin}\left(2\mathrm{x}\right)-\mathrm{Csinx}-4\mathrm{Dcos}\left(2\mathrm{x}\right)-\left[-\mathrm{Asinx}+2\mathrm{Bcos}\left(2\mathrm{x}\right)+\mathrm{Ccosx}-2\mathrm{Dsin}\left(2\mathrm{x}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\left[\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)\right]=\mathrm{cosx}-\mathrm{sin}2\mathrm{x}\\ ⇒\left(-3\mathrm{A}-\mathrm{C}\right)\mathrm{cosx}+\left(-3\mathrm{C}+\mathrm{A}\right)\mathrm{sinx}+\left(-6\mathrm{D}-2\mathrm{B}\right)\mathrm{cos}\left(2\mathrm{x}\right)+\left(-6\mathrm{B}+2\mathrm{D}\right)\mathrm{sin}\left(2\mathrm{x}\right)=\mathrm{cosx}-\mathrm{sin}2\mathrm{x}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}-3\mathrm{A}-\mathrm{C}=1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(3\right)\\ -6\mathrm{B}+2\mathrm{D}=-1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(4\right)\\ -3\mathrm{C}+\mathrm{A}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(5\right)\\ -6\mathrm{D}-2\mathrm{B}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(6\right)\end{array}$

Solve the equation (3) and (5),

role="math" localid="1655096234005" $\begin{array}{c}3\left(-3\mathrm{A}-\mathrm{C}\right)=1×3\\ -3\mathrm{C}-9\mathrm{A}=3\\ \text{\hspace{0.17em}}-3\mathrm{C}+\mathrm{A}=0\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{A}=\frac{-3}{10}\end{array}$

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

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

Solve the equation (4) and (6),

role="math" localid="1655096203374" $3\left(-6\mathrm{B}+2\mathrm{D}\right)=-1×3\phantom{\rule{0ex}{0ex}}\begin{array}{c}6\mathrm{D}-18\mathrm{B}=-3\\ -6\mathrm{D}-2\mathrm{B}=0\\ \mathrm{B}=\frac{3}{20}\end{array}$

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

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

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{Acosx}+\mathrm{Bsin}\left(2\mathrm{x}\right)+\mathrm{Csinx}+\mathrm{Dcos}\left(2\mathrm{x}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\mathrm{sin}\left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\mathrm{cos}\left(2\mathrm{x}\right)\end{array}$

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

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)\\ \mathrm{y}={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{x}}-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\mathrm{sin}\left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\mathrm{cos}\left(2\mathrm{x}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(7\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=\frac{-7}{20},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\text{'}\left(0\right)=\frac{1}{5}$

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

$\begin{array}{c}\frac{-7}{20}={\mathrm{c}}_{1}{\mathrm{e}}^{2\left(0\right)}+{\mathrm{c}}_{2}{\mathrm{e}}^{-0}-\frac{3}{10}\mathrm{cos}\left(0\right)+\frac{3}{20}\mathrm{sin}\left(0\right)-\frac{1}{10}\mathrm{sin}\left(0\right)-\frac{1}{20}\mathrm{cos}\left(0\right)\\ \frac{-7}{2}={\mathrm{c}}_{1}+{\mathrm{c}}_{2}-\frac{3}{10}-\frac{1}{20}\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=-\frac{7}{20}+\frac{3}{10}+\frac{1}{20}\\ {\mathrm{c}}_{1}+{\mathrm{c}}_{2}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(8\right)\end{array}$

Now find the derivative of the equation (7),

$\mathrm{y}\text{'}=2{\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{x}}-{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{x}}+\frac{3}{10}\mathrm{sinx}+\frac{3}{10}\mathrm{cos}\left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{cosx}+\frac{1}{10}\mathrm{sin}\left(2\mathrm{x}\right)$

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

$\begin{array}{c}\frac{1}{5}=2{\mathrm{c}}_{1}{\mathrm{e}}^{2\left(0\right)}-{\mathrm{c}}_{2}{\mathrm{e}}^{-0}+\frac{3}{10}\mathrm{sin}\left(0\right)+\frac{3}{10}\mathrm{cos}\left(0\right)-\frac{1}{10}\mathrm{cos}\left(0\right)+\frac{1}{10}\mathrm{sin}\left(0\right)\\ \frac{1}{5}=2{\mathrm{c}}_{1}-{\mathrm{c}}_{2}+\frac{3}{10}-\frac{1}{10}\\ 2{\mathrm{c}}_{1}-{\mathrm{c}}_{2}=\frac{1}{5}-\frac{3}{10}+\frac{1}{10}\\ 2{\mathrm{c}}_{1}-{\mathrm{c}}_{2}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(9\right)\end{array}$

Solve the equation (8) and (9),

$\begin{array}{c}{\mathrm{c}}_{1}+{\mathrm{c}}_{2}=0\\ 2{\mathrm{c}}_{1}-{\mathrm{c}}_{2}=0\\ {c}_{1}=0\end{array}$

Substitute the value of ${\mathrm{C}}_{1}=0$ in the equation (8),

$\begin{array}{c}{\mathrm{c}}_{1}+{\mathrm{c}}_{2}=0\\ {\mathrm{c}}_{2}=0\end{array}$

Substitute the value of ${\mathrm{c}}_{1}=0$ and ${\mathrm{c}}_{2}=0$ in the equation (7),

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

Thus, the initial solution to the differential equation is:

$\mathrm{y}=-\frac{3}{10}\mathrm{cosx}+\frac{3}{20}\mathrm{sin}\left(2\mathrm{x}\right)-\frac{1}{10}\mathrm{sinx}-\frac{1}{20}\mathrm{cos}\left(2\mathrm{x}\right)$ ### Want to see more solutions like these? 