For what values of r does the function y =

Mary Jackson 2021-12-19 Answered
For what values of r does the function y=erx satisfy the differential equation y4y+y=0?
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Expert Answer

Vivian Soares
Answered 2021-12-20 Author has 36 answers
Step 1
1) Find y and y
2) Substitute the expressions for y, y, y in the given differential equation
3) Solve for r
Step 2
It is given that
y=erx
Differentiate with respect to x
y=d(erx)dx
The Chain Rule for differential
d[f[g(x)]]dx=d[f[g(x)]]d[g(x)]×d[g(x)]dx
dydx=d(erx)d(rx)×d(rx)dx
y=erx×(r×1×x11)
Remember that:
1) the derivative of ex is ex itself
2) the derivative of xn is nxn1, this is known as power rule
y=rerx
Step 3
Note that:
y=ry
Deffirntiate again, to get
y=ry=r×rerx=r2erx
Substitute the expressions for y, y, y in the following differential equation
y4y+y=0
r2erx4rerx+erx=0
Since erx is never 0, we can divide both sides by it
r24r+1=0
add 3 to both sides
r24r+4=3
Notice that the LHS is a perfect square
(r2)2=3
Take the square root of both sides
r2=±3
Add 2 to both sides
r=2±3

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Thomas White
Answered 2021-12-21 Author has 40 answers
Step 1
Derivatives.
y=rerx
y=r2erx
Step 2
Substitute into differential equation
r2erx4rerx+erx=0
Step 3
Factor
erx(r24r+1)=0
Step 4
Solve:
erx=0 or r24r+1=0
erx=0 does not exist
r24r+1=0 must be solved with quadratic formula.
Step 5
Quadratic formula:
r=4±(4)24×1×12×1
r=4±1642
r=2±122
r=2±3

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nick1337
Answered 2021-12-27 Author has 575 answers

Step 1
y=1erx
y=rerx
y=r2erx
r24r+1=0
r=[4±164]2
r=[4±23]2
r=2±3

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