Question

Solve the integral equation $\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}{\mathbf{e}}^{\mathbf{-}\mathbf{\tau }}\mathbf{f}\mathbf{\left(}\mathit{\tau }\mathbf{\right)}\mathit{d}\mathit{\tau }$.

Short answer

The required solution for the integral equation$f\left(t\right)=e^t+e^t{\int }_0^t{e}^{-\tau }f\left(\tau \right)d\tau$is $f\left(t\right)=e^{2t}$.

Step-by-step solution

Define convolution:

Consider two functionsandthat are continuous on the interval$[0,\infty )$, then the convolution of the two functionsandcan be denoted as$\mathbf{f}\mathbf{*}\mathbf{g}$.

$\mathbf{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{*}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}*{\int }_0^t\mathbf{f}\left(\tau \right)\mathbf{g}\mathbf{(}\mathbf{t}-\tau \mathbf{)}\mathbf{d}\tau$

Solve the equation by using Laplace transform:

Consider the given equation,

$f\left(t\right)=e^t+e^t{\int }_0^t{e}^{-\tau }f\left(\tau \right)d\tau$

Apply Laplace transform on both sides,

$\mathcal{L}\left\{e^t\right\}+\mathcal{L}\left\{{\int }_0^t{e}^{t-\tau }f\left(\tau \right)\text{d}\tau \right\}=\mathcal{L}\left\{f\right(t\left)\right\}\left(\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}\right)$

$\frac{1}{s-1}+\mathcal{L}\left\{{\int }_0^t{e}^{t-\tau }f\left(\tau \right)\text{d}\tau \right\}=\mathcal{L}\left\{f\right(t\left)\right\}$$--->\left(1\right)$

Hence, to find$\mathcal{L}\left\{{\int }_0^t{e}^{t-\tau }f\left(\tau \right)\text{d}\tau \right\}$, use the convolution formula and Laplace transform,

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$=\mathcal{L}\left\{e^t\right\}\mathcal{L}\left\{f\right(t\left)\right\}\left(\mathcal{L}\left\{e^{at}\right\}=\frac{1}{s-a}\right)$

$=\frac{1}{s-1}·\mathcal{L}\left\{f\right(t\left)\right\}---\left(2\right)$

Evaluate the equation by substituting the values:

Thus, substitute equationin,

$\frac{1}{s-1}+\frac{1}{s-1}·\mathcal{L}\left\{f\right(t\left)\right\}=\mathcal{L}\left\{f\right(t\left)\right\}$

$\frac{1}{s-1}·\mathcal{L}\left\{f\right(t\left)\right\}-\mathcal{L}\left\{f\right(t\left)\right\}=-\frac{1}{s-1}$

$\left(\frac{1}{s-1}-1\right)\mathcal{L}\left\{f\right(t\left)\right\}=-\frac{1}{s-1}$

$\left(\frac{1-s+1}{s-1}\right)\mathcal{L}\left\{f\right(t\left)\right\}=-\frac{1}{s-1}$

$\left(\frac{2-s}{s-1}\right)\mathcal{L}\left\{f\right(t\left)\right\}=-\frac{1}{s-1}$