Determine the differentiable function $f$ whose graph lies above the x-axis and passes through the point $(0, 1)$

Question

I'm trying to solve the question:

Determine the differentiable function $f$ whose graph lies above the $x$-axis and passes through the point $(0, 1)$ and such that for any $x\geq0$ the area under the graph of $y = f(x)$ above the interval $[0, x]$ is four times the slope of the tangent line to the graph of $y = f(x)$ at the point $(x, f(x))$.

I have written $$\int_0^xf(x)dx=4f'(x)$$ to start with, but I'm not sure if this is correct.

Any help would be appreciated. Thanks in advance.

Answer 1

Clearly since $\int_0^xf(t)dt=4f'(x)$ for all $x\in\mathbb{R}$, we have that $f'$ is differentiable and, $$\frac d{dx}\int_0^xf(t)dt=f(x)=4f''(x)$$

Note that,

The general solution of the differential equation $4y''-y=0$ is $y=A\sinh(\frac x2)+B\cosh(\frac x2)$ for abitrary $A,B\in\mathbb{R}$.

Hence $f(x)=A\sinh(\frac x2)+B\cosh(\frac x2)$ for all $x\in\mathbb{R}$ for some $A,B\in\mathbb{R}$ and since $f(0)=1$ we have that $B=1$. Also $\int_0^0f(x)dx=4f'(0)$gives $f'(0)=\frac12A=0$. Therefore the only solution to the equation is $f:\mathbb{R}\to\mathbb{R}$ defined by, $$f(x)=\cosh\left(\frac x2\right)\text{ for all }x\in\mathbb{R}$$

Answer 2

Since$$\int_0^xf(t)\,\mathrm dt=4f'(x),\tag1$$differentiating both sides you get that $f(x)=4f''(x)$. So, $f(x)=a\cosh\left(\frac x2\right)+b\sinh\left(\frac x2\right)$ for some constants $a$ and $b$. Since $f(0)=1$, $a=1$. On the other hand, it follows from $(1)$ that $f'(0)=0$. So, $b=0$, and $f(x)=\cosh\left(\frac x2\right)$.