Key Concepts and Formulas

• Leibniz Integral Rule: Used to differentiate an integral with variable limits. For $\frac{d}{dx} \int_{a(x)}^{b(x)} g(t, x) dt = g(b(x), x) b'(x) - g(a(x), x) a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x} g(t, x) dt$. In our case, $g(t,x)$ does not depend on $x$, so the last term is zero.
• First-Order Linear Differential Equations / Separation of Variables: Techniques to solve differential equations of the form $\frac{dy}{dx} = P(x)Q(y)$.
• Logarithm Properties: Including $\ln(e^a) = a$ and $e^{\ln a} = a$.

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Step-by-Step Solution

Step 1: Differentiate the Integral Equation We are given the integral equation: $f(x) = \int\limits_0^x {{e^t}f(t)dt + {e^x}} \quad \text{.... (1)}$ To eliminate the integral and obtain a differential equation, we differentiate both sides of equation (1) with respect to $x$. $\frac{d}{dx} f(x) = \frac{d}{dx} \left( \int\limits_0^x {{e^t}f(t)dt} \right) + \frac{d}{dx} \left( {e^x} \right)$ Applying the Leibniz Integral Rule to the integral term, where $g(t) = e^t f(t)$, the lower limit $a(x) = 0$, and the upper limit $b(x) = x$: $\frac{d}{dx} \left( \int\limits_0^x {{e^t}f(t)dt} \right) = e^x f(x) \cdot \frac{d}{dx}(x) - e^0 f(0) \cdot \frac{d}{dx}(0)$ Since $\frac{d}{dx}(x) = 1$ and $\frac{d}{dx}(0) = 0$, this simplifies to $e^x f(x)$. The derivative of $e^x$ is $e^x$. Thus, the differentiated equation is: $f'(x) = e^x f(x) + e^x$

Step 2: Rearrange into a Separable Differential Equation The differential equation obtained is $f'(x) = e^x f(x) + e^x$. We can factor out $e^x$ from the right-hand side: $f'(x) = e^x (f(x) + 1)$ This is a first-order differential equation that can be solved using the method of separation of variables. We rearrange it so that all terms involving $f(x)$ and $f'(x)$ are on one side, and all terms involving $x$ are on the other. Dividing both sides by $(f(x) + 1)$ (assuming $f(x) + 1 \neq 0$): $\frac{f'(x)}{f(x) + 1} = e^x$

Step 3: Integrate Both Sides Now, we integrate both sides with respect to $x$: $\int \frac{f'(x)}{f(x) + 1} dx = \int e^x dx$ For the left-hand side, let $u = f(x) + 1$. Then, $du = f'(x) dx$. The integral becomes $\int \frac{1}{u} du = \ln|u| = \ln|f(x) + 1|$. The right-hand side integral is $\int e^x dx = e^x$. Equating the results and adding a constant of integration, $C$: $\ln|f(x) + 1| = e^x + C$

Step 4: Determine the Constant of Integration To find the value of $C$, we use the original integral equation (1) to find an initial condition. Substitute $x=0$ into equation (1): $f(0) = \int\limits_0^0 {{e^t}f(t)dt + {e^0}}$ The definite integral from $0$ to $0$ is $0$, and $e^0 = 1$. So, $f(0) = 0 + 1 = 1$. Now, substitute $x=0$ and $f(0)=1$ into the integrated equation $\ln|f(x) + 1| = e^x + C$: $\ln|1 + 1| = e^0 + C$ $\ln(2) = 1 + C$ Solving for $C$: $C = \ln(2) - 1$

Step 5: Solve for $f(x)$ Substitute the value of $C$ back into the integrated equation: $\ln|f(x) + 1| = e^x + (\ln(2) - 1)$ To remove the absolute value, we observe from the original equation that $f(x) = \int_0^x e^t f(t) dt + e^x$. For small positive $x$, $f(x)$ is dominated by $e^x$, which is positive. It can be shown that $f(x)+1 > 0$ for all $x$. $\ln(f(x) + 1) = e^x + \ln(2) - 1$ We can rewrite the right side using logarithm properties: $\ln(f(x) + 1) = e^x - 1 + \ln(2)$ $\ln(f(x) + 1) = \ln(e^{e^x - 1}) + \ln(2)$ $\ln(f(x) + 1) = \ln(2 \cdot e^{e^x - 1})$ Exponentiating both sides with base $e$: $f(x) + 1 = 2 \cdot e^{e^x - 1}$ Finally, solve for $f(x)$: $f(x) = 2 \cdot e^{e^x - 1} - 1$

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always remember to add the constant $C$ when integrating. This constant is essential for finding the particular solution.
• Incorrect Application of Leibniz Rule: Ensure the integrand is a function of the integration variable ($t$) and the limits are functions of the differentiation variable ($x$).
• Errors in Logarithm/Exponentiation: Be careful when manipulating logarithmic and exponential terms; use properties correctly to simplify.

Summary

The problem involves solving an integral equation by converting it into a differential equation. We differentiated the given equation using the Leibniz Integral Rule to obtain a first-order differential equation. This differential equation was then solved using the method of separation of variables. An initial condition was found by substituting $x=0$ into the original integral equation. This initial condition allowed us to determine the constant of integration, leading to the explicit form of $f(x)$.

The final answer is $f(x) = 2 \cdot e^{e^x - 1} - 1$. This corresponds to option (C).

The final answer is $\boxed{2{e^{{e^x} - 1}} - 1}$.