0

Let $f(x)$ twice differentiable. Show that $$ f''(x) = f(x),\ f(0) = 1,\ f'(0) = 0 \quad\Longrightarrow\quad f'(x)\cosh x - f(x)\sinh x = c_0 \in \Bbb R $$

I had tried to interchange each hyperbolics with exponentials but cannot find way to represent it into a certain constant in $\Bbb R$

Any hint?

Björn Friedrich
  • 3,747
Beverlie
  • 2,645

2 Answers2

1

Let $F(x)=f'(x)\cosh x-f(x)\sinh x$. Then

$$\begin{align} F'(x) &=(f''(x)\cosh x+f'(x)\sinh x)-(f'(x)\sinh x+f(x)\cosh h)\\ &=(f''(x)-f(x))\cosh x\\ &=0 \end{align}$$

since $f''(x)=f(x)$. This implies $F(x)$ is constant. From

$$F(0)=f'(0)\cosh0-f(0)\sinh0=0\cdot1-1\cdot0=0$$

we have $f'(x)\cosh x-f(x)\sinh x=0$ for all $x$.

Barry Cipra
  • 79,832
0

Start with:

$$f''(x) = f(x)$$

Multiply both sides with $\cosh x$:

$$f''(x) \cosh x = f(x) \cosh x$$

Transform the left side:

$$(f'(x) \cosh x)' - f'(x) \sinh x = f(x) \cosh x$$

From here it follows:

$$(f'(x) \cosh x)' = f'(x) \sinh x + f(x) \cosh x$$

or:

$$(f'(x) \cosh x)' = (f(x) \sinh x)'$$

$$(f'(x) \cosh x - f(x) \sinh x)' = 0$$

which means that

$$f'(x) \cosh x - f(x) \sinh x = c_0 \in \Bbb R $$

Boundary conditions $f(0),f'(0)$ simply don't matter.

Saša
  • 15,906