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Separable differential equation |
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| Apr25-06, 04:37 PM | #1 |
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Separable differential equation
okay... i got this problem
sovle the separable differential equation 4x-2y(x^2+1)^(1/2)(dy/dx)=0 using the following intial condition: y(0) = -3 y^2 = ? (function of x) I guess that means the constant is -3 so i put all the x on 1 side and all the y on one side 4x = 2y(x^2+1)^(1/2)(dy/dx) (4x)(dx) = 2y(x^2+1)^(1/2)(dy) (4xdx)/(x^2+1)^(1/2) = 2ydy integral both sides I got 4(x^2+1)^(1/2) = y^2 i tried the following answers y^2 = 4(x^2+1)^(1/2) y^2 = 4(x^2+1)^(1/2)+9 y^2 = 4(x^2+1)^(1/2)-3 they are all wrong!!! WHAT IS WRONG?! IS MY WAY OF DOING IT TATALLY WRONG?! |
| Apr25-06, 05:36 PM | #2 |
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This is correct. [tex]y^2=4\sqrt{x^2+1}+C[/tex]
Now plug in your initial condition to solve for C. |
| Apr25-06, 05:40 PM | #3 |
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well you got most of it but i dont know why you are trying 9 and -3 as c.
it says y(0) = -3 y = +-4*(x^2+1) + c so y(0) = +-(0^2+1) + c = -3 can you figure it out from here |
| Apr25-06, 06:08 PM | #4 |
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Separable differential equation
if
y^2 = 4(x^2+1)^(1/2)+c y = sqrt(4(x^2+1)^(1/2)+c) c = 5 is the correct answer. THANX!!! |
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