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.