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### 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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{=}}{\mathbf{6}}{\mathbf{t}}{\mathbf{;}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{3}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{0}\right){\mathbf{=}}{\mathbf{-}}{\mathbf{1}}$

The solution to the initial value problem is $\mathrm{y}={\mathrm{t}}^{3}-\mathrm{t}+3$.

See the step by step solution

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

The differential equation is,

$\mathrm{y}\text{'}\text{'}=6\mathrm{t}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

${\mathrm{y}}^{\text{'}\text{'}}=0$

The auxiliary equation for the above equation,

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

## Step 2: Now find the complementary solution of the given equation.

The root of an auxiliary equation is,

${\mathrm{m}}_{1}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=0$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}\mathrm{t}+{\mathrm{c}}_{2}$

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

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{2}\left(\mathrm{At}+\mathrm{B}\right)\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{At}}^{3}+{\mathrm{Bt}}^{2}\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=3{\mathrm{At}}^{2}+2\mathrm{Bt}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=6\mathrm{At}+2\mathrm{B}\end{array}$

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

$\begin{array}{c}\mathrm{y}\text{'}\text{'}=6\mathrm{t}\\ 6\mathrm{At}+2\mathrm{B}=6\mathrm{t}\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}6\mathrm{A}=6\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=1\\ 2\mathrm{B}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{B}=0\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{2}\left(\mathrm{At}+\mathrm{B}\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{2}\left(\left(1\right)\mathrm{t}+0\right)\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{3}\end{array}$

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

Therefore, the general solution is,

$\begin{array}{l}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_{1}\mathrm{t}+{\mathrm{c}}_{2}+{\mathrm{t}}^{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(3\right)\end{array}$

Given the initial condition,

$\mathrm{y}\left(0\right)=3,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{y}\text{'}\left(0\right)=-1$

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

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_{1}\mathrm{t}+{\mathrm{c}}_{2}+{\mathrm{t}}^{3}\\ 3={\mathrm{c}}_{1}\left(0\right)+{\mathrm{c}}_{2}+0\\ {\mathrm{c}}_{2}=3\end{array}$

Now find the derivative of the equation (3),

$\mathrm{y}\text{'}={\mathrm{c}}_{1}+3{\mathrm{t}}^{2}$

Substitute the value of y’ = -1 and t = 0 in the above equation,

$\begin{array}{c}-1={\mathrm{c}}_{1}+3{\left(0\right)}^{2}\\ {\mathrm{c}}_{1}=-1\end{array}$

Substitute the value of ${\mathrm{c}}_{1}=-1$ and ${\mathrm{c}}_{2}=3$ in the equation (3), we get:

$\begin{array}{c}\mathrm{y}={\mathrm{c}}_{1}\mathrm{t}+{\mathrm{c}}_{2}+{\mathrm{t}}^{3}\\ \mathrm{y}=\left(-1\right)\mathrm{t}+3+{\mathrm{t}}^{3}\\ \mathrm{y}={\mathrm{t}}^{3}-\mathrm{t}+3\end{array}$

Thus, the general solution is $\mathrm{y}={\mathrm{t}}^{3}-\mathrm{t}+3.$