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Differential Equation & Linear Algebra Help?

  • Thread starter JSGhost
  • Start date
  • #1
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Homework Statement


Solve the following differential equation
8x - 2y sqrt(x^2 + 1) dy/dx = 0
subject to the initial condition: y(0) = -3.

y = ?

Homework Equations



Separable DE's
dy/dx = g(x)/f(y)

The Attempt at a Solution



8x - 2y sqrt(x^2 + 1) dy/dx = 0
8x = 2y sqrt(x^2 + 1) dy/dx
[8x/sqrt(x^2+1)]dx = 2y dy
integrate both sides

u = x^2 + 1
du = 2x dx

4 (integral) [1/sqrt(u)] du = y^2
4(2u^(1/2)) + c = y^2
8 sqrt(x^2+1) + c = y^2

Since y(0) = -3.
Substitute to find c.
8sqrt((0)^2+1) + c = (-3)^2
8 + c = 9
c = 1

y^2 = 8 sqrt(x^2+1) + 1

I get that answer but it isn't correct. I am using webwork and need help solving the equation. Thanks.
 

Answers and Replies

  • #2
tiny-tim
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Hi JSGhost! :smile:

(have a square-root; √ and try using the X2 tag just above the Reply box :wink:)
Solve the following differential equation
8x - 2y sqrt(x^2 + 1) dy/dx = 0
subject to the initial condition: y(0) = -3.

y^2 = 8 sqrt(x^2+1) + 1
Looks ok so far …

what was your final answer, "y = … " ? :smile:
 
  • #3
26
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y^2 = 8 sqrt(x^2 + 1) + 1

to

y = sqrt( 8(x^2+1)^(1/2) + 1) ?

I've tried that already.
 
  • #4
tiny-tim
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Hi JSGhost! :smile:

(just got up :zzz: …)
y = sqrt( 8(x^2+1)^(1/2) + 1) ?
("sqrt" and "1/2" ? and what happened to that √ I gave you? :wink:)

erm :redface:

that doesn't look negative to me! :biggrin:
 
  • #5
26
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Nothing seems to work. Maybe I am approaching the problem wrong? Maybe it's not separable...but linear first-order equation.

y = sqrt( 8sqrt(x^1 + 1) + 1) doesn't work. Those symbols don't work cause I can't copy those into the answer placement.
 
  • #6
tiny-tim
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If you put x = 0 into sqrt( 8sqrt(x^1 + 1) + 1), you get √(8√(1) + 1) = √(8+1) = 3,

and you want -3

sooo … ? :smile:
 
  • #7
HallsofIvy
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If [itex]y^2= f(x)[/itex] y is NOT necessarily equal to [itex]\sqrt{f(x)}[/itex]! There are two possible solutions and only one can satisfy y(0)= -3.
 
  • #8
26
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Thanks. Finally got it. I needed a negative. :cry: I thought sqrt(9) would give negative and positive of 3.

y = - sqrt( 8sqrt(x^1 + 1) + 1)
 
  • #9
tiny-tim
Science Advisor
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he he:biggrin:

a maths symbol only ever has its unique "principal value" …

see http://en.wikipedia.org/wiki/Principal_value" [Broken] :wink:
 
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