Question

Solve \(y^{(4)} - 8y''' + 26y'' - 40y' + 25y =0 \)

Answer 1

As you should know in differential equations, it has been proven in away a general solution for a DE, and the best option would be a function that its derivative would be the same, and that would be an exponential function, of the form e^wx... let me go through the steps

Answer 2

let \[y(x) = e^{\omega x}\]

As you know that the nth derivative of y(x) should be...

\[\frac{d^n}{dx^n}(e^{\omega x}) = \omega^n e^{\omega x}\]

So that means the first derivative is:

\[\frac{d e^{(\omega x )}}{dx}= \omega e^{\omega x}\]

the 2nd derivative yields to:

\[\frac{d^2 e^{(\omega x )}}{dx^2}= \omega^2 e^{\omega x}\]

the 3rd derivative results in:

\[\frac{d^3 e^{(\omega x )}}{dx^3}= \omega^3 e^{\omega x}\]

and the 4th derivative implies:

\[\frac{d^4 e^{(\omega x )}}{dx^4}= \omega^4 e^{\omega x}\]

Answer 3

\[\frac{d^4y}{dx^4}-8\frac{d^3y}{dx^3}+26 \frac{d^2y}{dx^2}-40\frac{dy}{dx} +25y =0\]

Answer 4

\[\omega^4 e^{\omega x}-8 \omega^3 e^{\omega x}+26 \omega^2 e^{\omega x}-40 \omega e^{\omega x}+25e^{\omega x} =0\]

Answer 5

let's somplify the formula by eliminating e^wx, that gives us

\[e^{\omega x}(\omega^4-8 \omega^3+26 \omega^2 -40 \omega+25) =0\]

hence,

\[ \omega^4-8 \omega^3+26 \omega^2 -40 \omega+25 =0\]

Answer 6

factorizing the above 4th order polynomial :

\[( \omega ^2 -4 \omega +5)^2=0\]

Now we need to find the roots of this equation using the quadratic formula:

\[\omega = 2 \pm i \space or \space \omega = 2-i \space or 2-i\]

now we plug our roots into the satisfying solution we get

\[y_1(x) =C_1e^{(2+i)x}\]
\[y_2(x) =C_2e^{(2-i)x}\]
\[y_3(x) =C_3e^{(2+i)x}\]
\[y_4(x) =C_4e^{(2+i)x}\]

Answer 7

\[y(x) = y_1(x)+y_2(x)+y_3(x)+y_4(x)\]

Answer 8

\[ =C_1e^{(2+i)x}+C_2e^{(2-i)x}+C_3e^{(2-i)x}+C_4e^{(2-+i)x}\]

Answer 9

please note the solution include 2 roots of 2+i and 2 roots of 2-i

Answer 10

Now let's apply one of the mose important formulas in mathematics, and that would be the Euler's formula which indicates:

\[e^{\xi+i \psi}= e^{\xi}\cos(\psi)+ie^{\xi}\sin(\psi)\]

Answer 11

now i think the rest is simple you are only one step away from the solution can you try it and if not let me know i will help you with the rest, ask any question you want... all the best