Question

In Problems 9–14 use the Laplace transform to solve the given initial value problem. Use the table of Laplace transforms in Appendix C as needed

${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}$

Short answer

Solution of given initial value problem is$\mathrm{y}=2\cos 3\mathrm{t}+\frac{5}{3}\sin 3\mathrm{t}+\frac{1}{6}\mathrm{tsin}3\mathrm{t}$

Step-by-step solution

Laplace transformation of a linear differential equation with constant coefficient

1. The Laplace transformation of a linear differential equation with constant coefficient is express a.${\mathbf{a}}_n\mathbf{L}\left\{\frac{d^{n}y}{{\mathrm{dt}}^n}\right\}\mathbf{=}{\mathbf{a}}_n\left[{\mathbf{s}}^n\mathbf{Y}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{-}{\mathbf{s}}^{(n-1)}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{-}\mathbf{L}\mathbf{-}{\mathbf{y}}^{(n-1)}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\right]$.

2. The formula for the inverse Laplace transform is as follows,

$\begin{array}{l}{\mathbf{L}}^{-1}\left\{\frac{}{s^2+k^2}\right\}\mathbf{=}\mathbf{sinkt}\\ {\mathbf{L}}^{-1}\left\{\frac{}{s^2+k^2}\right\}\mathbf{=}\mathbf{coskt}\\ {\mathbf{L}}^{-1}\left\{\frac{2\mathrm{ks}}{{(s^2+k^2)}^2}\right\}\mathbf{=}\mathbf{tsinkt}\end{array}$

3. The formula for the Laplace transform of $\cos kt$is as follows,

$\mathbf{L}\mathbf{\left\{}\mathbf{coskt}\mathbf{\right\}}\mathbf{=}\frac{s}{{\mathbf{s}}^2\mathbf{+}{\mathbf{k}}^2}$

Apply the Laplace transform of the given linear differential equation

Use the above definition to write the Laplace transform of the given linear differential equation

$s^{2}L\left\{y\right\}-sy\left(0\right)-y^{\text{'}}\left(0\right)+9L\left\{y\right\}=L\left\{\cos 3t\right\}$

Put the value y (0) =2 and y' (0) =5 in above expression.

$\begin{array}{l}\left(s^2+9\right)L\left\{y\right\}-2s-5=\frac{s}{s^2+9}\left(s^2+9\right)\\ L\left\{y\right\}=\frac{s}{s^2+9}+2s+5\left(s^2+9\right)\\ L\left\{y\right\}=\frac{2s^3+5s^2+19s+45}{\left(s^2+9\right)}\\ L\left\{y\right\}=\frac{2s^3+5s^2+19s+45}{{\left(s^2+9\right)}^2}\end{array}$

From the above expression,

localid="1668588681080" $L\left\{y\right\}=\frac{2s^3+5s^2+19s+45}{{\left(s^2+9\right)}^2}..........1$

Evaluate the value of a0 and a1

Simplify Equation (1) using partial fractions as follows

$\frac{2{\mathrm{s}}^3+5{\mathrm{s}}^2+19\mathrm{s}+45}{{({\mathrm{s}}^2+9)}^2}=\frac{{\mathrm{a}}_1\mathrm{s}+{\mathrm{a}}_0}{{\mathrm{s}}^2+9}+\frac{{\mathrm{a}}_3\mathrm{s}+{\mathrm{a}}_2}{{({\mathrm{s}}^2+9)}^2}$

From the above expression,

$\frac{2{\mathrm{s}}^3+5{\mathrm{s}}^2+19\mathrm{s}+45}{{({\mathrm{s}}^2+9)}^2}=\frac{{\mathrm{a}}_1\mathrm{s}+{\mathrm{a}}_0}{{\mathrm{s}}^2+9}+\frac{{\mathrm{a}}_3\mathrm{s}+{\mathrm{a}}_2}{{({\mathrm{s}}^2+9)}^2}.$-------(2)

Multiply with${({\mathrm{s}}^2+9)}^2$ on both sides of equation (2).

$\frac{(2{\mathrm{s}}^3+5{\mathrm{s}}^2+19\mathrm{s}+45){({\mathrm{s}}^2+9)}^2}{{({\mathrm{s}}^2+9)}^2}=\frac{({\mathrm{a}}_1\mathrm{s}+{\mathrm{a}}_0){({\mathrm{s}}^2+9)}^2}{{\mathrm{s}}^2+9}+\frac{({\mathrm{a}}_3\mathrm{s}+{\mathrm{a}}_2){({\mathrm{s}}^2+9)}^2}{{({\mathrm{s}}^2+9)}^2}$

$2{\mathrm{s}}^3+5{\mathrm{s}}^2+19\mathrm{s}+45={\mathrm{a}}_1{\mathrm{s}}^3+{\mathrm{a}}_0{\mathrm{s}}^2+\mathrm{s}(9{\mathrm{a}}_1+{\mathrm{a}}_3)+(9{\mathrm{a}}_0+{\mathrm{a}}_2)$

Comparing the constant terms in above expression as follows.

Evaluate the coefficients of s2 and s

Simplifying the above equations for${\mathrm{a}}_2$ by comparing the coefficients of$s^2$ as follows.

Simplifying the above equations for$a_3$ by comparing the coefficients of $s$ as follows.

Substitute the value of$a_0,a_1,a_2$ and$a_3$ in equation (2).

$\frac{2s^3+5s^2+19s-45}{{(s^2+9)}^2}=\frac{s}{{(s^2+9)}^2}+\frac{2s+5}{s^2+9}$

Above expression is simplified as follows,

$\begin{array}{c}\frac{2{\mathrm{s}}^3+5{\mathrm{s}}^2+19\mathrm{s}-45}{{({\mathrm{s}}^2+9)}^2}=\frac{\mathrm{s}}{{({\mathrm{s}}^2+9)}^2}+\frac{2\mathrm{s}+5}{{\mathrm{s}}^2+9}\\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{2\mathrm{s}}{{\mathrm{s}}^2+9}+\frac{5}{{\mathrm{s}}^2+9}+\frac{\mathrm{s}}{{({\mathrm{s}}^2+9)}^2}\end{array}$

From the above expression,

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{2\mathrm{s}}{{\mathrm{s}}^2+9}+\frac{5}{{\mathrm{s}}^2+9}+\frac{\mathrm{s}}{{({\mathrm{s}}^2+9)}^2}.$----(3)

Apply inverse Laplace transform

Take inverse Laplace on both sides of equation (3),

Therefore, the solution of given initial value problem is $\mathrm{y}=2\cos 3\mathrm{t}+\frac{5}{3}\sin 3\mathrm{t}+\frac{1}{6}\mathrm{tsin}3\mathrm{t}$.