Question

Use the Laplace transform to solve the given integral equation or integraodifferential equation. $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{2}\underset{\mathbf{0}}{\overset{\mathbf{t}}{\int }}\mathbf{f}\mathbf{\left(}\mathbf{\tau }\mathbf{\right)}\mathbf{cos}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{\tau }\mathbf{)}\mathbf{d\tau }\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{+}\mathbf{sint}$

Short answer

Using the Laplace transformation we get_{$f\left(t\right)=\left[4-7t+4t^2\right]e^{-t}$}

Step-by-step solution

We will use the Convolution Theorem;

Solution of integrodifferential equation of using Laplace equation.

_{$L\left\{\underset{0}{\overset{t}{\int }}f\left(\tau \right)g\left(t-\tau \right)d\tau \right\}=L\left\{f\left(t\right)\right\}L\left\{g\left(t\right)\right\}$}

Solving further;

Apply Laplace operator on both side of expression and by using Laplace linearity property we get;

_{$\begin{array}{l}f\left(t\right)+2\underset{0}{\overset{t}{\int }}f\left(\tau \right)\cos (t-\tau )d\tau =4e^{-t}+\sin t\\ L\left\{f\left(t\right)\right\}+2L\left\{\underset{0}{\overset{t}{\int }}f\left(\tau \right)\cos (t-\tau )d\tau =4e^{-t}\right\}=4L\left\{e^{-t}\right\}+L\left\{\sin t\right\}\\ \\ F\left(s\right)+\frac{2F\left(s\right)s}{s^2+1}=\frac{4}{s+1}+\frac{1}{s^2+1}\\ F\left(s\right)=4\frac{s^2+1}{{\left(s+1\right)}^3}+\frac{1}{{\left(s+1\right)}^2}\end{array}$}

Using Partial Differentiation:

Evaluating constant coefficient by usual method we get A=4 B=-8 C=8, Now find function value f(t) by using inverse Laplace transform

_{$4\frac{s^2+1}{{\left(s+1\right)}^3}=\frac{A}{s+1}+\frac{B}{{\left(s+1\right)}^2}+\frac{C}{{\left(s+1\right)}^3}$}

Putting A=4,B=-8 and C=8.Now, using the formulas of inverse transformation;

$\begin{array}{l}{L}^{-1}\left\{F\left(s\right)\right\}=4L^{-1}\left\{\frac{1}{s+1}\right\}-8L^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}+8L^{-1}\left\{\frac{1}{{\left(s+1\right)}^3}\right\}+L^{-1}\left\{\frac{1}{{\left(s+1\right)}^2}\right\}\\ f\left(t\right)=4e^{-t}+-8t{e}^{-t}+4t^2{e}^{-t}+t{e}^{-t}\\ f\left(t\right)=\left[4-7t+4t^2\right]e^{-t}\end{array}$

Hence, the final answer is:$f\left(t\right)=\left[4-7t+4t^2\right]e^{-t}$