Question 74

Let $$f(x) = \int_0^x e^t f(t)dt + e^x$$ be a differentiable function for all $$x \in R$$. Then $$f(x)$$ equals:

Solution

$$f(x)=\int_0^x e^t f(t)\,dt+e^x$$

Putting

$$x=0,$$

$$f(0)=0+1=1$$

Differentiate both sides:

$$f'(x)=e^x f(x)+e^x$$

$$f'(x)=e^x(f(x)+1)$$

Let

$$y=f(x)$$

Then,

$$\frac{dy}{dx}=e^x(y+1)$$

Separating variables,

$$\frac{dy}{y+1}=e^x\,dx$$

Integrating,

$$\ln(y+1)=e^x+C$$

Using

$$y(0)=1,$$

$$\ln2=1+C$$

$$C=\ln2-1$$

Hence,

$$\ln(y+1)=e^x+\ln2-1$$

Therefore,

$$y+1=e^{e^x+\ln2-1}$$

$$=2e^{e^x-1}$$

Thus,

$$f(x)=2e^{e^x-1}-1$$

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Let $$f(x) = \int_0^x e^t f(t)dt + e^x$$ be a differentiable function for all $$x \in R$$. Then $$f(x)$$ equals:

Solution

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