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Expert-verified Found in: Page 180 ### 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 a particular solution to the differential equation.${\mathbf{2}}{\mathbf{x}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{x}}{\mathbf{=}}{\mathbf{3}}{{\mathbf{t}}}^{2}$

The particular solution is ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^{2}-12\mathrm{t}+24.$

See the step by step solution

## Step 1: Use the method of undetermined coefficients to find a particular solution to the differential equation.

Consider the given differential equation,

$2\mathrm{x}\text{'}+\mathrm{x}=3{\mathrm{t}}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

According to the method of undetermined coefficients, the particular solution of the differential equation;

$\mathrm{ax}\text{'}\text{'}+\mathrm{bx}\text{'}+\mathrm{cx}={\mathrm{dt}}^{\mathrm{m}},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{m}=0,1,2,3,...$

It is of the form ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+{\mathrm{A}}_{\mathrm{m}-1}{\mathrm{t}}^{\mathrm{m}-1}+...+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}$

Comparing the above equation with equation (1),

We get, m = 2

## Step 2: Find a particular solution to the differential equation for m = 2

Therefore, the particular solution of equation (1),

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_{2}{\mathrm{t}}^{2}+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

Now find the derivative of above equation,

${\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2{\mathrm{A}}_{2}\mathrm{t}+{\mathrm{A}}_{1}$

From the equation (1), substitute the value of ${\mathrm{x}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)$, we get

$\begin{array}{c}2{\mathrm{x}}_{\mathrm{p}}\text{'}+{\mathrm{x}}_{\mathrm{p}}=3{\mathrm{t}}^{2}\\ 2\left[2{\mathrm{A}}_{2}\mathrm{t}+{\mathrm{A}}_{1}\right]+{\mathrm{A}}_{2}{\mathrm{t}}^{2}+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}=3{\mathrm{t}}^{2}\\ 4{\mathrm{A}}_{2}\mathrm{t}+2{\mathrm{A}}_{1}+{\mathrm{A}}_{2}{\mathrm{t}}^{2}+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}=3{\mathrm{t}}^{2}\\ {\mathrm{A}}_{2}{\mathrm{t}}^{2}+\left[4{\mathrm{A}}_{2}+{\mathrm{A}}_{1}\right]\mathrm{t}+2{\mathrm{A}}_{1}+{\mathrm{A}}_{0}=3{\mathrm{t}}^{2}\end{array}$

## Step 3: Final conclusion.

Comparing the all coefficients of the above equation,

$\begin{array}{c}{\mathrm{A}}_{2}=3\\ 4{\mathrm{A}}_{2}+{\mathrm{A}}_{1}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(3\right)\\ 2{\mathrm{A}}_{1}+{\mathrm{A}}_{0}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(4\right)\end{array}$

Substitute the value of ${\mathrm{A}}_{2}$ in the equation (3),

$\begin{array}{c}4\left(3\right)+{\mathrm{A}}_{1}=0\\ {\mathrm{A}}_{1}=-12\end{array}$

Substitute the value of ${\mathrm{A}}_{1}$ in the equation (4),

$\begin{array}{c}2\left(-12\right)+{\mathrm{A}}_{0}=0\\ {\mathrm{A}}_{0}=24\end{array}$

Substitute the value of ${\mathrm{A}}_{0},\text{\hspace{0.17em}}{\mathrm{A}}_{1}$and ${\mathrm{A}}_{2}$ in the equation (2),

${\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^{2}-12\mathrm{t}+24$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_{2}{\mathrm{t}}^{2}+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}\\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^{2}+\left(-12\right)\mathrm{t}+24\\ {\mathrm{x}}_{\mathrm{p}}\left(\mathrm{t}\right)=3{\mathrm{t}}^{2}-12\mathrm{t}+24\end{array}$ ### Want to see more solutions like these? 