# What is the general solution of the differential equation? : d^4y/dx^4−4y=0

Leonel Schwartz 2022-09-29 Answered
What is the general solution of the differential equation? : $\frac{{d}^{4}y}{{dx}^{4}}-4y=0$
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## Answers (1)

Bleha7s
Answered 2022-09-30 Author has 11 answers
With linear operator $D=\frac{d}{dx}$, this is:
$\left({D}^{4}-4\right)y=0$
$\left({D}^{2}-2\right)\left({D}^{2}+2\right)y=0$
$\left(D-\sqrt{2}\right)\left(D+\sqrt{2}\right)\left(D-i\sqrt{2}\right)\left(D+i\sqrt{2}\right)y=0$
Each of these factors has a simple solution, eg:
1. $\left(D-\sqrt{2}\right)y=0⇒y\prime =\sqrt{2}y⇒y=\alpha {e}^{\sqrt{2}x}$
2. $\left(D-i\sqrt{2}\right)y=0⇒y\prime =i\sqrt{2}y⇒y=\beta {e}^{i\sqrt{2}x}$
The complete solution is the superposition of individual solutions:
$⇒y\left(x\right)=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$
$C{e}^{i\sqrt{2}x}+D{e}^{-i\sqrt{2}x}$ can also be re-written using the Euler formula , arriving at:
$y=A{e}^{\sqrt{2}x}+B{e}^{-\sqrt{2}x}+C\prime \mathrm{cos}\left(\sqrt{2}x\right)+D\prime \mathrm{sin}\left(\sqrt{2}x\right)$
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