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

_{${\mathbf{y}}^{\text{'}\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}^{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}$}

Where$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\left\{\begin{array}{l}1,0\le t<\frac{\pi }{2}\\ sint,t\ge \frac{\pi }{2}\end{array}\right\}$

Short answer

Solution of given initial value problem is$1-(1-\sin t)U\left(t-\frac{\pi }{2}\right)-\frac{1}{2}\left(t-\frac{\pi }{2}\right)\cos tU\left(t-\frac{\pi }{2}\right).$

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,

role="math" localid="1668591049965" $\begin{array}{l}{\mathbf{L}}^{-1}\mathbf{\left\{}\frac{\mathbf{k}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}\mathbf{\right\}}\mathbf{=}\mathbf{sinkt}\\ {\mathbf{L}}^{-1}\mathbf{\left\{}\frac{\mathbf{s}}{{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}}\mathbf{\right\}}\mathbf{=}\mathbf{coskt}\\ {\mathbf{L}}^{-1}\mathbf{\left\{}\frac{\mathbf{2}\mathbf{ks}}{{\mathbf{(}{\mathbf{s}}^{\mathbf{2}}\mathbf{+}{\mathbf{k}}^{\mathbf{2}}\mathbf{)}}^{\mathbf{2}}}\mathbf{\right\}}\mathbf{=}\mathbf{tsinkt}\end{array}$

3. The formula for the Laplace transform of 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.

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

Use the above definition to write the function $\mathrm{f}\left(\mathrm{t}\right)$ as follows,

$\mathrm{f}\left(\mathrm{t}\right)=1-\mathrm{U}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)+\mathrm{sintU}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)$

Put the value $y\left(0\right)=1$and$y^{\text{'}}\left(0\right)=0$ in above expression.

$\begin{array}{c}({\mathrm{s}}^2+1)\mathrm{L}\left\{\mathrm{y}\right\}-\mathrm{s}=\mathrm{L}\left\{1-\mathrm{U}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)+\mathrm{sintU}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)\right\}\\ ({\mathrm{s}}^2+1)\mathrm{L}\left\{\mathrm{y}\right\}=\mathrm{s}+\mathrm{L}\left\{1\right\}-\mathrm{L}\left\{\mathrm{U}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)\right\}+\mathrm{L}\left\{\mathrm{sintU}\left(\mathrm{t}-\frac{\mathrm{\pi }}{2}\right)\right\}\\ ({\mathrm{s}}^2+1)\mathrm{L}\left\{\mathrm{y}\right\}=\mathrm{s}+\frac{1}{\mathrm{s}}-\frac{{\mathrm{e}}^{-\frac{\mathrm{\pi s}}{2}}}{\mathrm{s}}+{\mathrm{e}}^{-\frac{\mathrm{\pi \pi }}{2}}\mathrm{L}\left\{\mathrm{cost}\right\}\\ ({\mathrm{s}}^2+1)\mathrm{L}\left\{\mathrm{y}\right\}=\frac{{\mathrm{s}}^2+1}{\mathrm{s}}-\frac{{\mathrm{e}}^{-\frac{\mathrm{\pi s}}{2}}}{\mathrm{s}}+\frac{{\mathrm{se}}^{-\frac{\mathrm{\pi x}}{2}}}{{\mathrm{s}}^2+1}\end{array}$

Above expression is simplified as,

$\begin{array}{c}({\mathrm{s}}^2+1)\mathrm{L}\left\{\mathrm{y}\right\}=\frac{{\mathrm{s}}^2+1}{\mathrm{s}}-\frac{{\mathrm{e}}^{-\frac{\mathrm{\pi s}}{2}}}{\mathrm{s}}+\frac{{\mathrm{se}}^{-\frac{\mathrm{\pi s}}{2}}}{{\mathrm{s}}^2+1}\\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{\mathrm{s}}-{\mathrm{e}}^{-\frac{\mathrm{\pi s}}{2}}\left\{\frac{{\mathrm{s}}^2+1-{\mathrm{s}}^2}{\mathrm{s}({\mathrm{s}}^2+1)}\right\}+\frac{{\mathrm{se}}^{-\frac{\mathrm{\pi s}}{2}}}{{({\mathrm{s}}^2+1)}^2}\\ \mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{\mathrm{s}}-{\mathrm{e}}^{-\frac{\mathrm{\pi s}}{2}}\left\{\frac{{\mathrm{s}}^2+1}{\mathrm{s}({\mathrm{s}}^2+1)}-\frac{\mathrm{s}}{({\mathrm{s}}^2+1)}\right\}+\frac{{\mathrm{se}}^{-\frac{\mathrm{\pi s}}{2}}}{{({\mathrm{s}}^2+1)}^2}\end{array}$

From the above expression,

$\mathrm{L}\left\{\mathrm{y}\right\}=\frac{1}{\mathrm{s}}-{\mathrm{e}}^{-\frac{\mathrm{\pi x}}{2}}\left\{\frac{{\mathrm{s}}^2+1}{\mathrm{s}({\mathrm{s}}^2+1)}-\frac{\mathrm{s}}{({\mathrm{s}}^2+1)}\right\}+\frac{{\mathrm{se}}^{-\frac{\mathrm{\pi x}}{2}}}{{({\mathrm{s}}^2+1)}^2}----\left(1\right)$

Evaluate the coefficients of s2

Above expression is simplify as follows,

$\begin{array}{l}L\left\{y\right\}=\frac{1}{s}-e^{-\frac{\pi s}{2}}\left\{\frac{s^2+1}{s\left(s^2+1\right)}-\frac{s}{\left(s^2+1\right)}\right\}+\frac{s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}\\ L\left\{y\right\}=\frac{1}{s}-e^{-\frac{\pi s}{2}}\left\{\frac{1}{s}-\frac{s}{\left(s^2+1\right)}\right\}+\frac{s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}\end{array}$

From the above expression,

$L\left\{y\right\}=\frac{1}{s}-e^{-\frac{\pi s}{2}}\left\{\frac{1}{s}-\frac{s}{\left(s^2+1\right)}\right\}+\frac{s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}----\left(2\right).$

Apply inverse Laplace transform

Take inverse Laplace on both sides of equation ( 2 ),

$\begin{array}{l}y=L^{-1}\left\{\frac{1}{s}-e^{-\frac{\pi s}{2}}\left\{\frac{1}{s}-\frac{s}{\left(s^2+1\right)}\right\}+\frac{1}{2}\frac{2s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}\right\}\\ y=L^{-1}\left\{\frac{1}{s}-e^{-\frac{\pi s}{2}}\left\{\frac{1}{s}-\frac{s}{\left(s^2+1\right)}\right\}+\frac{1}{2}\frac{2s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}\right\}\\ =L^{-1}\left\{\frac{1}{s}\right\}-L^{-1}\left\{\frac{e^{-\frac{\pi s}{2}}}{s}\right\}+L^{-1}\left\{\frac{s{e}^{-\frac{\pi s}{2}}}{\left(s^2+1\right)}\right\}+\frac{1}{2}{L}^{-1}\left\{\frac{2s{e}^{-\frac{\pi s}{2}}}{{\left(s^2+1\right)}^2}\right\}\\ =1-U\left(t-\frac{\pi }{2}\right)+\cos \left(t-\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)+\frac{1}{2}\left(t-\frac{\pi }{2}\right)\sin \left(t-\frac{\pi }{2}\right)U\left(t-\frac{\pi }{2}\right)\\ =1-U\left(t-\frac{\pi }{2}\right)+\sin tU\left(t-\frac{\pi }{2}\right)+\frac{1}{2}\left(t-\frac{\pi }{2}\right)\cos tU\left(t-\frac{\pi }{2}\right)\\ =1-(1-\sin t)U\left(t-\frac{\pi }{2}\right)-\frac{1}{2}\left(t-\frac{\pi }{2}\right)\cos tU\left(t-\frac{\pi }{2}\right)\end{array}$

Therefore, the solution of given initial value problem is

$1-(1-\sin t)U\left(t-\frac{\pi }{2}\right)-\frac{1}{2}\left(t-\frac{\pi }{2}\right)\cos tU\left(t-\frac{\pi }{2}\right).$