Differential equation

**Rapha** · December 25th 2008, 12:40 AM

Hello everyone! (Merry Christmas and happy holidays)

I got to solve

-y'' + y = 0

I'm so lost.
I tried cos(x) and sin(x)

then i tried y = sin(-x) => y' = -cos(-x) => y'' = -sin(-x)

=> $-y'' + y = -(-sin(-x))+ sin(-x) = sin(-x)+sin(-x) = -2 sin(x) \not= 0$

The solution is not that obvious, I guess.

Does anyone knows the solution?

Kind regards,
Rapha

**thelostchild** · December 25th 2008, 01:53 AM

The DE can be rearranged to the form

$\frac{d^2y}{dx^2}=y$

what is a function that when differentiated twice gives the function itself

Spoiler:

you can also solve this by separating variables twice

**mr fantastic** · December 25th 2008, 02:41 AM

Originally Posted by Rapha

Hello everyone! (Merry Christmas and happy holidays)

I got to solve

-y'' + y = 0

I'm so lost.
I tried cos(x) and sin(x)

then i tried y = sin(-x) => y' = -cos(-x) => y'' = -sin(-x)

=> $-y'' + y = -(-sin(-x))+ sin(-x) = sin(-x)+sin(-x) = -2 sin(x) \not= 0$

The solution is not that obvious, I guess.

Does anyone knows the solution?

Kind regards,
Rapha

The DE has the general form $a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0$ .

The standard technique is to try a solution of the form $y = A e^{\lambda x}$ .

Read these:

PlanetMath: second order linear differential equation with constant coefficients

Homogeneous 2nd-order differential equations

to get started.

**Soroban** · December 25th 2008, 05:33 AM

Hello, Rapha!

This is one of the most basic forms . . .

Solve: . $-y'' + y \:=\:0$

We have: . $y'' - y \:=\:0$

Let: $y \:=\:e^{mx}\quad\Rightarrow\quad y' \:=\:me^{mx}\quad\Rightarrow\quad y'' \:=\:m^2e^{mx}$

. . Substitute: . $m^2e^{mx} - e^{mx} \:=\:0$

. . Divide by $e^{mx}\!:\;\;m^2 - 1 \:=\:0 \quad\Rightarrow\quad m \:=\:\pm1$

. . Hence, the solutions are: . $y \:=\:e^x,\;\;y \:=\:e^{-x}$

Form a linear combination of the solutions: . $y \;=\;C_1e^x + C_2e^{-x}$