1. ## Differential equation

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)

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

The solution is not that obvious, I guess.

Does anyone knows the solution?

Kind regards,
Rapha

2. The DE can be rearranged to the form

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

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

Spoiler:

$\displaystyle y=e^x$

you can also solve this by separating variables twice

3. 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)

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

The solution is not that obvious, I guess.

Does anyone knows the solution?

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

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

PlanetMath: second order linear differential equation with constant coefficients

Homogeneous 2nd-order differential equations

to get started.

4. Hello, Rapha!

This is one of the most basic forms . . .

Solve: .$\displaystyle -y'' + y \:=\:0$
We have: .$\displaystyle y'' - y \:=\:0$

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

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

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

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

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

5. y''-y=0

The characteristic equation is:$\displaystyle y^2-1=0$

$\displaystyle y_1=-1$
$\displaystyle y_2=1$

The solution is $\displaystyle y=C1*e^{x*y_1}+C2*e^{x*y_2}$

$\displaystyle y=C1*e^{x}+C2*e^{-x}$