Question

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

Answer 1

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 linear differential equation: y⁽⁴⁾- y = 0

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

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

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

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

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

Where: $D_1 = C_3 + C_4 and D_2= i ( C_3 - C_4 )$

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

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