# Need to check the method (DE)

#### idontknow

Equation : $$\displaystyle yâ€™â€™=y$$.
Add both sides +yâ€™ then substitution as $$\displaystyle yâ€™+y=t$$.
$$\displaystyle tâ€™=t$$.
Is this way correct?

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#### romsek

Math Team
The usual way is to form the characteristic equation.

$y''-y=0$

$s^2 - 1= 0$

$s = \pm 1$

$y = c_1 e^t + c_2 e^{-t}$

2 people

#### Joppy

Also the chant "what when differentiated twice equals itself!"

1 person

#### romsek

Math Team
Also the chant "what when differentiated twice equals itself!"
I don't remember hearing that one down the pub.

#### skipjack

Forum Staff
Is this way correct?
Yes. Having found $t$, you could then solve $y' + y = t$ by using an integrating factor.

The same integrating factor works for the original equation.

Integrating $e^xy'' - e^xy = 0$ gives $e^xy' - e^xy = \text{A}$, etc.

2 people

#### topsquark

Math Team
I don't remember hearing that one down the pub.
Apparently you go to the wrong bars. :dance:

-Dan

1 person