 Bobbie Comstock

2021-12-31

Find the solution of the following Second Order Differential Equations.
$4y4{y}^{\prime }+y=0$ reinosodairyshm

Expert

Calculation:
Convert $4y4{y}^{\prime }+y=0$ into characteristics equation.
$4{D}^{2}+4D+1=0$
${\left(2D\right)}^{2}+2\cdot 2D\cdot 1+{1}^{2}=0$
${\left(2D+1\right)}^{2}=0$
$D=-\frac{1}{2}$
Thus, the solution of differential equation SlabydouluS62

Expert

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.
$x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
$m=\frac{-4±\sqrt{{4}^{2}-4\left(1\right)\left(4\right)}}{2\left(4\right)}$
$m=\frac{-4±\sqrt{0}}{8}$
$m=-\frac{4}{8}-0,-\frac{4}{8}+0$
$m=-\frac{1}{2},-\frac{1}{2}$
Since values of m are equal, then the solution of the given differential equation is:
$y=\left({c}_{1}+{c}_{2}x\right){e}^{-\frac{1}{2}x}$
Where, ${C}_{1},{C}_{2}$ are arbitrary constant. karton

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