Vikolers6

2021-12-27

Find the general solution to the following differential equations.

$y6{y}^{\prime}+13y=0$

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$y={e}^{\lambda t}$

$\left({e}^{\lambda t}\right)6{\left({e}^{\lambda t}\right)}^{\prime}+13{e}^{\lambda t}=0$

${e}^{\lambda t}({y}^{2}+6y+13)=0$

${y}^{2}+6y+13=0$

Roots of the quadratic equation in the form$a{x}^{2}+bx+c=0$ is

$x}_{1,2}=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a$

Here,$a=1,b=6\text{and}c=13$

$y}_{1,2}=\frac{-6\pm \sqrt{{6}^{2}-4\times 1\times 13}}{2\times 1$

$y}_{1,2}=\frac{-6\pm \sqrt{36-52}}{2$

${y}_{1,2}=-3-2i,-3+2i$

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:$y={e}^{\alpha t}({c}_{1}\mathrm{cos}\left(\beta t\right)+{c}_{2}\mathrm{sin}\left(\beta t\right))$

$y={e}^{-3t}({c}_{1}\mathrm{cos}\left(2t\right)+{c}_{2}\mathrm{sin}\left(2t\right))$

A second order linear, homogenous ODE has the form

For an equation

Rewrite the equation with

Roots of the quadratic equation in the form

Here,

For two complex roots

So, the general solution takes the form:

xandir307dc

Beginner2021-12-29Added 35 answers

Answer:

karton

Expert2022-01-10Added 613 answers

We were given the next equation

The roots of the characteristic equation

are

The general solution of Eq. (1) is, by the Theorem 4.3.2, given by

where

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inverse laplace transform - with symbolic variables:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{({s}^{2}-{a}^{2})(s-2b)}$

My steps:

$F(s)=\frac{2{s}^{2}+(a-6b)s+{a}^{2}-4ab}{(s+a)(s-a)(s-2b)}$

$=\frac{A}{s+a}+\frac{B}{s-a}+\frac{C}{s-2b}+K$

$K=0$

$A=F(s)\ast (s+a)$