Find the solution of the following Second Order Differential Equations. 4y"+4y'+y=0

Bobbie Comstock

Bobbie Comstock

Answered question

2021-12-31

Find the solution of the following Second Order Differential Equations.
4y4y+y=0

Answer & Explanation

reinosodairyshm

reinosodairyshm

Beginner2022-01-01Added 36 answers

Calculation:
Convert 4y4y+y=0 into characteristics equation.
4D2+4D+1=0
(2D)2+22D1+12=0
(2D+1)2=0
D=12
Thus, the solution of differential equation 4y4y+y=0 is y=Aex2+Bxex2
SlabydouluS62

SlabydouluS62

Skilled2022-01-02Added 52 answers

The auxiliary equation of this equation is:
m2+4m+m=0
On comparing this equation with the general quadratic equation,
ax2+bx+c=0
Then, a=1,b=4,c=1
Now, use the quadratic form to find the value of m.
x=b±b24ac2a
m=4±424(1)(4)2(4)
m=4±08
m=480,48+0
m=12,12
Since values of m are equal, then the solution of the given differential equation is:
y=(c1+c2x)e12x
Where, C1,C2 are arbitrary constant.
karton

karton

Expert2022-01-09Added 613 answers

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

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