Solving 2nd order homogenous linear ODE with a squared coefficient

Question

I am attempting to solve a differential equation of the form:

$$ \frac{d^2 f}{d x^2} - \gamma^2 f = 0$$

I have set up and solved the characteristic equation as:

$$ \begin{align} z^2 - \gamma^2 z + 0z = 0 \\ z (z - \gamma^2 )= 0 \end{align}$$

which is satisfied when $z=0$ and when $z = \gamma^2$

There are two distinct roots ($r_1$ and $r_2$) hence the general solution should be of the form

$$ f(x) = A e^{r_1 x} + Be^{r_2 x}$$

In this case:

$$ \begin{align} f(x) &= A e^{0 \times x} + Be^{\gamma^2 x} \\ &= A + Be^{\gamma^2 x} \end{align} $$

Apparently, this solution is incorrect as in the answers it is given as:

$$ f(x) = Ae^{\gamma x} + B e^{-\gamma x}$$

I would like to know where I went wrong to give me the incorrect solution

Answer 1

HINT

Without appealing directly to the characteristic equation, you can also proceed as follows \begin{align*} f'' - \gamma^{2}f = 0 & \Longleftrightarrow (f'' - \gamma f') + (\gamma f' - \gamma^{2}f) = 0\\\\ & \Longleftrightarrow (f' - \gamma f)' + \gamma(f' - \gamma f) = 0\\\\ & \Longleftrightarrow g' + \gamma g = 0 \end{align*}

where $g = f' - \gamma f$. Can you take it from here?

Answer 2

$\frac{\partial^2 f}{\partial x^2}-\gamma^2f=0$

By guessing a solution to be of the form of $f = e^{zx}$ and substituting it in the equation, we get

$$\frac{\partial^2}{\partial x^2}\left(e^{zx}\right)-\gamma^2\left(e^{zx}\right)=0$$

$$\Rightarrow z^2e^{zx}-\gamma^2e^{zx}=0$$

$$\Rightarrow e^{zx}\left(z^2-\gamma^2\right)=0$$

For non-trivial solution ($z\neq -\infty$), we can have $e^{zx}\neq0$

$$\Rightarrow z^2-\gamma^2=\frac{0}{e^{zx}}=0$$

The above equation is the characteristic equation.

Now, by solving the characteristic equation, we get

$$z = \pm \gamma$$

By putting these in the guessed solution, we get $f = e^{\gamma x}$ and $f = e^{-\gamma x}$.

Now, these are linearly independent solutions, the general solution of the differential equation would be any linear combination of these two solutions.

$$\Rightarrow f = A e^{\gamma x} + B e^{-\gamma x}$$.

Answer 3

You can solve it as follows:

$$ \frac{d^2 f}{d x^2} - \gamma^2 f = 0\leftrightarrow \left(\frac{d}{dx}+\gamma\right)\left(\frac{d}{dx}-\gamma\right)f $$

so calling

$$ \cases{ \left(\frac{d}{dx}+\gamma\right)u=0\\ u=\left(\frac{d}{dx}-\gamma\right)f } $$

we have

$$ \cases{ u = c_1 e^{\gamma x}\\ \left(\frac{d}{dx}-\gamma\right)f = c_1 e^{-\gamma x}\to f = c_2 e^{\gamma x}-\frac{c_1 e^{-\gamma x}}{2 \gamma } = c_2 e^{\gamma x}+c_3 e^{-\gamma x} } $$