Question

f(x) is a differentiable functionsuch that f(x)=x^2+integral 0tox e^-t f(x-t)dt then find f(x)

Answer 1

f(x)=x^2+\frac{x^3}{3}

Answer 2

Solution:

We are given

$$f(x)=x^2+\int_0^x e^{-t}f(x-t)\,dt.$$

1. Take Laplace transform:

Let $\mathcal{L}\{f(x)\}=F(s)$. Recall:

• $\mathcal{L}\{x^2\}=\frac{2}{s^3}$,
• The Laplace transform of the convolution $\int_0^x e^{-t}f(x-t)\,dt$ equals $\mathcal{L}\{e^{-t}\}\cdot F(s)$ and $\mathcal{L}\{e^{-t}\}=\frac{1}{s+1}$.

Thus, taking the Laplace transform of both sides,

$$F(s)=\frac{2}{s^3}+\frac{1}{s+1}F(s).$$

2. Solve for $F(s)$:

$$F(s) - \frac{1}{s+1}F(s)=\frac{2}{s^3} \quad\Rightarrow\quad F(s)\left(1-\frac{1}{s+1}\right)=\frac{2}{s^3}.$$

Simplify the factor:

$$1-\frac{1}{s+1}=\frac{s}{s+1}.$$

Then,

$$F(s)\cdot\frac{s}{s+1}=\frac{2}{s^3}\quad\Rightarrow\quad F(s)=\frac{2}{s^3}\cdot\frac{s+1}{s}=\frac{2(s+1)}{s^4}.$$

3. Inverse Laplace transform:

Express $F(s)$ as:

$$F(s)=\frac{2s}{s^4}+\frac{2}{s^4}=\frac{2}{s^3}+\frac{2}{s^4}.$$

Using known transforms:

• $\mathcal{L}^{-1}\left\{\frac{2}{s^3}\right\}=x^2$ (since $\mathcal{L}\{x^2\}=\frac{2}{s^3}$),
• $\mathcal{L}^{-1}\left\{\frac{2}{s^4}\right\}=\frac{x^3}{3}$ (since $\mathcal{L}\{x^3\}=\frac{6}{s^4}$).

Therefore,

$$f(x)=x^2+\frac{x^3}{3}.$$