Vikolers6

## Answered question

2021-12-27

Find the general solution to the following differential equations.
$y6{y}^{\prime }+13y=0$

### Answer & Explanation

alexandrebaud43

Beginner2021-12-28Added 36 answers

Given: $y6{y}^{\prime }+13y=0$
$y6{y}^{\prime }+13y=0$
A second order linear, homogenous ODE has the form $ayb{y}^{\prime }+cy=0$
For an equation $ayb{y}^{\prime }+cy=0$, assume a solution of the form ${e}^{\lambda t}$
Rewrite the equation with $y={e}^{\lambda t}$
$\left({e}^{\lambda t}\right)6{\left({e}^{\lambda t}\right)}^{\prime }+13{e}^{\lambda t}=0$
${e}^{\lambda t}\left({y}^{2}+6y+13\right)=0$
${y}^{2}+6y+13=0$
Roots of the quadratic equation in the form $a{x}^{2}+bx+c=0$ is
${x}_{1,2}=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
Here, $a=1,b=6\text{and}c=13$
${y}_{1,2}=\frac{-6±\sqrt{{6}^{2}-4×1×13}}{2×1}$
${y}_{1,2}=\frac{-6±\sqrt{36-52}}{2}$
${y}_{1,2}=-3-2i,-3+2i$
For two complex roots ${y}_{1}\ne {y}_{2}$, where ${y}_{1}=-3-2i,{y}_{2}=-3+2i$
So, the general solution takes the form: $y={e}^{\alpha t}\left({c}_{1}\mathrm{cos}\left(\beta t\right)+{c}_{2}\mathrm{sin}\left(\beta t\right)\right)$
$y={e}^{-3t}\left({c}_{1}\mathrm{cos}\left(2t\right)+{c}_{2}\mathrm{sin}\left(2t\right)\right)$

xandir307dc

Beginner2021-12-29Added 35 answers

$y={e}^{kx}$
${k}^{2}+6k+13=0$
$D={b}^{2}-4ac={6}^{2}-4\cdot 1\cdot 13=-16$
$\sqrt{D}=±4i$
${k}_{1,2}=-3±2i$
Answer: $y={e}^{-3x}\left({C}_{1}\mathrm{cos}2x+{C}_{2}\mathrm{sin}2x\right)$

karton

Expert2022-01-10Added 613 answers

We were given the next equation

The roots of the characteristic equation

are ${\lambda }_{1}=-3+2i,{\lambda }_{2}=-3-2i$
The general solution of Eq. (1) is, by the Theorem 4.3.2, given by
$y\left(t\right)={c}_{1}{e}^{\mu t}\mathrm{cos}vt+{c}_{2}{e}^{\mu t}\mathrm{sin}vt$
where ${\lambda }_{1,2}=\mu ±iv$. In this case, the solution is
$y\left(t\right)={c}_{1}{e}^{-3t}\mathrm{cos}2t+{c}_{2}{e}^{-3t}\mathrm{sin}2t$

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