Find $f(x)$ such that $2 \int_0^x f(t) \,\mathrm dt = x(f(x)+2000)$

Question

Let $f: \Bbb R \to \Bbb R$ be such that

$$2 \int_0^x f(t) \,\mathrm dt = x(f(x)+2000)$$

for every $x$. Find $f(x)$.

Answer 1

First "pull yourself by the hair" to prove that $f$ is differentiable: LHS is continuous in $x$. Therefore RHS is continuous in $x$, which means $f(x)$ is continuous everywhere except, maybe, $x=0$. This means LHS is differentiable everywhere except, maybe, $x=0$. Therefore the same is true for RHS, and also $f(x)$.

Now let $x \ne 0$ and equate the derivatives of both sides. This will give you a linear ODE of the first order which you can solve using basic ODE-solving tools.

Answer 2

Take the derivative of both sides with respect to $x$. Then by part 1 of the Fundamental Theorem of Calculus and Product Rule, we have: $$ 2f(x)=1\cdot(f(x)+2000)+x\cdot f'(x) \iff xf'(x)=f(x)-2000 $$ Let $y=f(x)$. Switching to Leibniz's Notation, we obtain a separable ODE: $$ \begin{align*} x\dfrac{dy}{dx} &= y-2000 \\ \int \dfrac{dy}{y-2000} &= \int \dfrac{dx}{x} \\ \ln|y-2000| &= \ln|x| +C \\ e^{\ln|y-2000|} &= e^{\ln|x|+C} \\ e^{\ln|y-2000|} &= e^Ce^{\ln|x|} \\ |y-2000| &= e^C|x| \\ y-2000 &= \pm e^Cx \\ y &= 2000+Kx \qquad (\text{where }K=\pm e^C)\\ \end{align*}$$

Hence, we have $\boxed{f(x)=2000+Kx}$, where $K$ is any real constant.

Answer 3

If the expressions on the left and right or equal, they still should be if you differentiate each side with respect to $x$. Doing this,

$$ 2 \ f(x) = \left(f(x) + 2000\right) + x \ f'(x) $$

Simplifying,

$$ f(x) = 2000 + x \ f'(x) $$

Rearranging and dividing through by $x$,

$$ f'(x) - \frac{1}{x}f(x) = -\frac{2000}{x} $$

This is a first-order ODE that can by solved by finding the integration factor $I$ as follows:

$$ I = \exp \left[ \int -\frac{1}{x} dx \right] \\ I = \exp \left[ -\ln\left|x\right| \right] = \frac{1}{x} $$

Solving the ODE,

$$ \left( \frac{1}{x}f(x) \right)' = -\frac{2000}{x^2} $$

Integrating both sides,

$$ \frac{1}{x}f(x) = \frac{2000}{x} + C \\ f(x) = 2000 + Cx $$

where $C$ is any real constant.

Answer 4

There is another approach, but a bit longer. We can use the Laplace Transformation here also. By taking L.T. from the both sides and knowing that:

• $\mathcal{L}\left(\int_0^x f(t)dt\right)=\frac{1}{s}\mathcal{L}(f(t))$

• $\mathcal{L}(tf(t))=-(\mathcal{L}(f(t))'$

we have $$2\frac{Y(s)}{s}+Y'(s)=\frac{2000}{s^2}$$ This is a linear OE with respect to $\mathcal{L}(f)=Y(s)$ which has a general solution as $Y(s)=\frac{2000s+C_1}{s^2}$. Now try to take Inverse L.T.to find the proper $f(t)$.