Expert Answer to "Find the general solution of the following differential..."

Brock Brown

Answered question

2021-12-31

Find the general solution of the following differential equation.
$y^{″} - 6 y^{'} + 13 y = 0$

raefx88y

Beginner2022-01-01Added 26 answers

We have to find the general solution of the following differential equation.

y 6 y^{'} + 13 y = 0

Let,

y = A e^{m x}

be the trial solution of (1)
Then

y^{'} = m A e^{m x}, y m^{2} A e^{m x}

The auxiliary equation becomes.

m^{2} - 6 m + 13 = 0

Using quadratic formula,

m = \frac{- (- 6) \pm \sqrt{{(- 6)}^{2} - 4 \cdot 1 \cdot 13}}{2 (1)}

\Rightarrow m = \frac{6 \pm \sqrt{36 - 52}}{2}

\Rightarrow m = \frac{6 \pm \sqrt{- 16}}{2}

m = \frac{6 \pm 4 i}{2}

m = 3 \pm 2 i

So, the general solution is

y = e^{3 x} C 4 \cos 2 x + c_{2} \sin 2 x, 4, 2

are arbitrary constants.

Bubich13

Beginner2022-01-02Added 36 answers

Get the equation in the form

C_{1} y^{n} + C_{2} y^{n - 1} + \dots + C_{n + 1} y = 0

y 6 y^{'} + 13 y = 0

Find the roots of the equation

C_{1} r^{n} + C_{2} r^{n - 1} + \dots + C_{n + 1}

r^{2} + 6 r + 13 = 0

r = - 3 \pm 2 i

Your result is

y = c_{1} e^{r 1 x} + c_{2} e^{r 2 x} + \dots + c_{n} e^{r n x}

y = e^{- 3 x} (c_{1} \cos (2 x) + c_{2} \sin (2 x))

karton

Expert2022-01-09Added 613 answers

y-6y+13y=0 We add all the numbers together, and all the variables 8y=0 y=0/8 y=0

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