Bobbie Comstock

2021-12-31

Find the solution of the following Second Order Differential Equations.

$4y4{y}^{\prime}+y=0$

reinosodairyshm

Beginner2022-01-01Added 36 answers

Calculation:

Convert$4y4{y}^{\prime}+y=0$ into characteristics equation.

$4{D}^{2}+4D+1=0$

${\left(2D\right)}^{2}+2\cdot 2D\cdot 1+{1}^{2}=0$

${(2D+1)}^{2}=0$

$D=-\frac{1}{2}$

Thus, the solution of differential equation$4y4y+y=0\text{}\text{is}\text{}y=A{e}^{-\frac{x}{2}}+Bx{e}^{-\frac{x}{2}}$

Convert

Thus, the solution of differential equation

SlabydouluS62

Skilled2022-01-02Added 52 answers

The auxiliary equation of this equation is:

${m}^{2}+4m+m=0$

On comparing this equation with the general quadratic equation,

$a{x}^{2}+bx+c=0$

Then,$a=1,b=4,c=1$

Now, use the quadratic form to find the value of m.

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

$m=\frac{-4\pm \sqrt{{4}^{2}-4\left(1\right)\left(4\right)}}{2\left(4\right)}$

$m=\frac{-4\pm \sqrt{0}}{8}$

$m=-\frac{4}{8}-0,-\frac{4}{8}+0$

$m=-\frac{1}{2},-\frac{1}{2}$

Since values of m are equal, then the solution of the given differential equation is:

$y=({c}_{1}+{c}_{2}x){e}^{-\frac{1}{2}x}$

Where,$C}_{1},{C}_{2$ are arbitrary constant.

On comparing this equation with the general quadratic equation,

Then,

Now, use the quadratic form to find the value of m.

Since values of m are equal, then the solution of the given differential equation is:

Where,

karton

Expert2022-01-09Added 613 answers

\(\begin{array}{} 4y+4y+y=0

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My steps:

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

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

$K=0$

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