# I How to solve the two following differential equations?

1. Nov 21, 2016

### Tspirit

(1) $\frac{d^{2}y}{dx^{2}}=0$
(2) $\frac{d^{2}y}{dx^{2}}=k^{2}y$, where k is a real positive number.

2. Nov 21, 2016

### haushofer

If the second derivative of a function vanishes, it can at most be linear in x. Same reasoning for your second equation, although looking at the notation I suspect you mean the rhs to have y(x) instead of x. If that is so, you want to try an exponential function like e^[kx].

3. Nov 21, 2016

### Tspirit

Thank you for your answer. But for the first equation, is there any process of deduction?

4. Nov 21, 2016

### Tspirit

Thank you very much for pointing the spelling error in the second equation, I have modified it.

5. Nov 21, 2016

### Tspirit

I have searched google for "harmonic oscillator", however, it is a little different to the equation I give. For harmonic oscillator, the equation is like this, $\frac{d^{2}y}{dx^{2}}=-k^{2}y$, where there is an additional negative sign.

6. Nov 21, 2016

### Tspirit

Now I know the solutions:
(1)For $$\frac{d^{2}y}{dx^{2}}=0,$$ the solution is $$y=A+Bx.$$
(2)For $$\frac{d^{2}y}{dx^{2}}=k^{2}y,$$ the solution is $$y=Ae^{-kx}+Be^{-kx}.$$
But I still don't know how they are deduced.

7. Nov 30, 2016

### haruspex

1) is easy. Just integrate both sides twice, remembering to plug in a constant of integration.
For 2), multiply both sides by dy/dx. Now you can integrate both sides. What do you get?

8. Nov 30, 2016

### LCKurtz

That is a constant coefficient linear DE of the form $y'' + \lambda y = 0$. The standard technique for such an equation is to assume a solution of the form $e^{rx}$. Plug that in and see what you get. You will want to consider three cases:$$\lambda = k^2 > 0,~\lambda = 0,~\lambda = -k^2 <0\text{.}$$

9. Nov 30, 2016

### haruspex

Sure, but I interpreted Tspirits' question as how to proceed if you have no idea what the answer is.

10. Dec 1, 2016

### lurflurf

$$\dfrac{d^2y}{{dx}^2}-k^2y=0$$
we have haruspex's sugestion
$$2\dfrac{d^2y}{{dx}^2}\dfrac{dy}{dx}-2k^2y\dfrac{dy}{dx}=0 \\ \dfrac{d}{dx}\left(\dfrac{dy}{dx}+k y\right)\left(\dfrac{dy}{dx}-k y\right)=0$$
we could factor the operator
$$\left(\dfrac{d}{dx}+k \right)\left(\dfrac{d}{dx}-k \right)y=0$$
we could solve for the inverse
$$-\left(\dfrac{dx}{dy}\right)^{-3}\dfrac{d^2x}{{dy}^2}-k^2y=0$$

11. Dec 1, 2016

### haruspex

I was thinking more along the lines of
y"=k2y
2y'y"=2k2yy'
y'2=k2y2+c2
Then some suitable hyperbolic trig substitution.