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enot [183]
3 years ago
13

Find the general solution to: y^(4) - y =0

Mathematics
1 answer:
N76 [4]3 years ago
4 0

Answer:

$y(t) = C_1 e^{t} + C_2 e^{-t} + D_1 \cos t + D_2 \sin t }$

Step-by-step explanation:

The equation is a<em> </em><em>linear differential equation: y⁽⁴⁾- y = 0 </em>

We assume the form of the solution y(t) is $y(t)=C_{1} e^{\alpha_{1} t} + C_{2} e^{\alpha_{2} t} + C_{3} e^{\alpha_{3} t} + C_{4} e^{\alpha_{4} t} $

where $\alpha_{i} are the roots of the auxiliary equation.

So, use the auxiliary equation: $\alpha^4 + 0 \alpha^3 + 0 \alpha^2 + 0 \alpha -1 =0$ to find the roots; the values are : α₁ = 1, α₂ = -1, α₃ = i, α₄ = -i

Then inserting $\alpha_{i} values in the assumed solution

⇒ <em>$y(t)=C_1 e^{t} + C_2 e^{-t} + C_{3} e^{it} + C_{4} e^{-it} $</em>

Also, because the last 2 terms have complex power, the solution can be written with cosine and sine terms:

<em>Using the Euler's formula: e^{ \pm i\theta } = \cos \theta \pm i\sin \theta, we can rewrite the solution as:</em>

$y(t) = C_{1} e^{t} + C_{2} e^{-t} + C_{3} e^{i t} + C_{4} e^{-i t}  = C_{1} e^{t} + C_{2} e^{-t} + C_{3} ( \cos t + i \sin t ) + C_{4} ( \cos t - i \sin t ) = C_{1} e^{t} + C_{2} e^{-t} + \cos t ( C_{3} + C_{4} ) + \sin t (i C_{3} - i C_{4} ) = C_{1} e^{t} + C_{2} e^{-t} + D_{1} \cos t +D_{2} \sin t$

<em>Where: </em>$D_1 = C_3 + C_4$ and $D_2= i ( C_3 - C_4 )$

<em>Finally the solution for de linear differential equation y^(4) - y =0 is:</em>

$y(t) = C_1 e^{t} + C_2 e^{-t} + D_1 \cos t + D_2 \sin t }$

<em> </em>

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Answer:

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Step-by-step explanation:

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