Question

(D^4+64) y=0 find the differential equation
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Answer 1

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Answer 2

SOLUTION

TO SOLVE

The differential equation

$\sf{ ({D}^{4} + 64)y = 0 }$

EVALUATION

Here the given differential equation is

$\sf{ ({D}^{4} + 64)y = 0 }$

Let $\sf{y = {e}^{mx} }$ be the trial solution

Then the auxiliary equation is

$\sf{ {m}^{4} + 64 = 0 }$

$\sf{ \implies \: {( {m}^{2} )}^{2} + {(8)}^{2} = 0 }$

$\sf{ \implies \: {( {m}^{2} + 8)}^{2} - 2. {m}^{2} .8 = 0 }$

$\sf{ \implies \: {( {m}^{2} + 8)}^{2} - 16 {m}^{2} = 0 }$

$\sf{ \implies \: {( {m}^{2} + 8)}^{2} - {(4m)}^{2} = 0 }$

$\sf{ \implies \: ( {m}^{2} + 8 + 4m) ( {m}^{2} + 8 - 4m) = 0 }$

$\sf{ \implies \: ( {m}^{2} + 4m + 8) ( {m}^{2} - 4m + 8) = 0 }$

Now $\sf{ ( {m}^{2} + 4m + 8) = 0 \: \: \: \: gives}$

$\displaystyle \sf{ \implies \: m = \frac{ - 4 \: \pm \: \sqrt{16 - 4.8} }{2.1} }$

$\displaystyle \sf{ \implies \: m = \frac{ - 4 \: \pm \: \sqrt{16 - 32} }{2} }$

$\displaystyle \sf{ \implies \: m = \frac{ - 4 \: \pm \: \sqrt{ - 16 } }{2} }$

$\displaystyle \sf{ \implies \: m = \frac{ - 4 \: \pm \: 4i }{2} }$

$\displaystyle \sf{ \implies \: m = - 2 \: \pm \: 2i }$

Similarly $\sf{ ( {m}^{2} - 4m + 8) = 0 \: \: \: \: gives}$

$\sf{ m = 2 \: \pm \: 2i}$

Thus four roots of the auxiliary equation are

- 2 + 2i , - 2 - 2i , 2 + 2i , 2 - 2i

Hence the required solution is

$\sf{y = {e}^{ - 2t} (a \cos 2t + b \sin 2t) + {e}^{ 2t} (c \cos 2t + d \sin 2t)}$

Where a , b , c , d are constants

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