Brock Brown

2021-12-31

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

raefx88y

Expert

We have to find the general solution of the following differential equation.
$y6{y}^{\prime }+13y=0$
Let, $y=A{e}^{mx}$ be the trial solution of (1)
Then ${y}^{\prime }=mA{e}^{mx},y{m}^{2}A{e}^{mx}$
The auxiliary equation becomes.
${m}^{2}-6m+13=0$
$m=\frac{-\left(-6\right)±\sqrt{{\left(-6\right)}^{2}-4\cdot 1\cdot 13}}{2\left(1\right)}$
$⇒m=\frac{6±\sqrt{36-52}}{2}$
$⇒m=\frac{6±\sqrt{-16}}{2}$
$m=\frac{6±4i}{2}$
$m=3±2i$
So, the general solution is
$y={e}^{3x}C4\mathrm{cos}2x+{c}_{2}\mathrm{sin}2x,4,2$ are arbitrary constants.

Bubich13

Expert

Get the equation in the form ${C}_{1}{y}^{n}+{C}_{2}{y}^{n-1}+\dots +{C}_{n+1}y=0$
$y6{y}^{\prime }+13y=0$
Find the roots of the equation ${C}_{1}{r}^{n}+{C}_{2}{r}^{n-1}+\dots +{C}_{n+1}$
${r}^{2}+6r+13=0$
$r=-3±2i$
Your result is $y={c}_{1}{e}^{r1x}+{c}_{2}{e}^{r2x}+\dots +{c}_{n}{e}^{rnx}$
$y={e}^{-3x}\left({c}_{1}\mathrm{cos}\left(2x\right)+{c}_{2}\mathrm{sin}\left(2x\right)\right)$

karton

Expert

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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