I'm at a loss about how to solve the following differential equation:
x'' = x^0.25
That is, double derivative of x equals x to the 0.25th power.
It's very frustrating because I have a feeling it's very trivial. Can anybody help, please? Thanks.
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Anonymous2007-09-02 11:24 ID:ZY7vOegH
0.25x?
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Anonymous2007-09-02 12:45 ID:D6QdaoP5
Differential equations work that way. x is a function, over a variable (y, t, or whatever). The solution will be x(y,c), that is, a function of the variable and an arbitrary constant. Well, two arbitrary constants, since it's a second derivative. I think.
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Anonymous2007-09-02 12:46 ID:D6QdaoP5
Meant to say "don't work that way".
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Anonymous2007-09-02 13:16 ID:c8mw+SRX
(16*x^(9/4)/45)
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Anonymous2007-09-02 13:22 ID:c8mw+SRX
int(x^0.25)=0.8x^1.25
int(0.8x^1.25)=(16*x^(9/4))/45
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Anonymous2007-09-02 13:35 ID:D6QdaoP5
No, no, no.
To check if a answer "f" is valid, you must:
1. Calculate "f" to the 0.25th power
2. Calculate the second derivative of "f"
3. The results from 1 and 2 must be equal
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Anonymous2007-09-02 16:29 ID:yr5AgAKM
constant of integration, anyone?
x'' = x^(1/4)
x' = (4/5)x^(5/4) + c
x = (4/5)(4/9)x^(9/4) + cx + d
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Anonymous2007-09-02 16:44 ID:vxqCpsDY
cant you jus revers solve this thing?
x'' = x ^ 0.25
x'' = (x^1/8)^2
x' = (1/3)(x^1/8)^3
x = (1/4)(1/3)(x^1/8)^3
if you think that the integral of (x^(1/8))^2 is (1/3)(x^(1/8))^3, then try differentiating the latter and see if you can get the former. and if you do, then you don't know how to differentiate, either.
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Anonymous2007-09-02 17:00 ID:vxqCpsDY
(1/3)(x^1/8)^3
power rule
3 * (1/3)(x^1/8)^(3-1)
(x^1/8)^2
No. x is a function which depends on a variable, t in this case.
To check a solution, you take its expression (such as "c1*sin(t)+c2*cos(t)") and substitute x by that. Then you substitute x' by its first derivative ("c1*cos(t)-c2*sin(t)" in this example). Then x'' by its second derivative ("-c1*sin(t)-c2*cos(t)" in this example).
So, you're not going to be able to solve explicitly for y I don't think, but you can give it a shot.
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4tran2007-09-02 21:52 ID:qxemZd8A
>>26
Mathematica dies if C1 =/= 0.
If C1 = 0, then y = [3(C +/- x)/Sqrt(40)]^(8/3)
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Anonymous2007-09-02 22:03 ID:bObOiDMP
Well, it's easy to do it if C1=0. If C1 != 0, then you can't solve for y explicitly. On top of that, I'm pretty confident that there's no closed form for Integral[(a+b*y^(5/4))^(-1/2),y], so you're not even going to be able to get that out.
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Anonymous2007-09-02 22:12 ID:bObOiDMP
And Mathematica should be able to solve it directly without inputing the solution form for the second order autonomous ODE.
DSolve[y''[x]-y[x]^(1/4)== 0,y,x] should do it for you.
>>30
I got 10 monstrous compounds of Beta and inverse Beta functions. I have no idea why they're relevent or why they're not outputted when doing the integral.
OK, that worked. What I did was use the BBCode tags to make it all pretty-like. Replace all curly brackets "{" and "}" in the following with straight brackets "[" and "]" to achieve the same result: