Need to check the method (DE)

Dec 2015
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Equation : \(\displaystyle y’’=y\).
Add both sides +y’ then substitution as \(\displaystyle y’+y=t\).
\(\displaystyle t’=t\).
Is this way correct?
 
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romsek

Math Team
Sep 2015
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The usual way is to form the characteristic equation.

$y''-y=0$

$s^2 - 1= 0$

$s = \pm 1$

$y = c_1 e^t + c_2 e^{-t}$
 
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Feb 2016
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Also the chant "what when differentiated twice equals itself!"
 
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skipjack

Forum Staff
Dec 2006
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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.
 
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