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Solution to nonlinear ODE with radicals |
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| Jul19-12, 12:44 AM | #1 |
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Solution to nonlinear ODE with radicals
I am not too familiar with differential equations but am familiar with basic calculus, I came across this equation trying to describe a particular function:
dy/dx =((sqrt((y-x)^2+y^2)-abs(y))/(y-x))*abs(y)/y Anyway I tried to separate the variables unsuccessfully and using v(x)=y(x)/x with no success, I really tried and couldn't solve it, so some insight would be appreciated. I tried restricting y to be positive, still without a breakthrough. Sorry if it's confusing, can't figure out the math format. |
| Jul19-12, 02:11 AM | #2 |
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There is a solution y = cx, if that helps. c satisfies a quartic. One of the roots is 1, but you can discount that, leaving a cubic. All the real roots of that are negative.
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| Jul19-12, 12:35 PM | #3 |
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Are you suggesting that I try to solve the differential equation with respect to c? Could you elaborate a little since I fail to see c satisfying a quartic.
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| Jul19-12, 05:24 PM | #4 |
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Recognitions:
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Solution to nonlinear ODE with radicals
Suppose y > 0 in some domain:
dy/dx = (√((y-x)2+y2)-y)/(y-x) Put y = cx: c = (√((c-1)2+c2)-c)/(c-1) c(c-1) + c = √((c-1)2+c2) = c2 (c-1)2+c2 = c4 (c-1)2 = c2(c2-1) There's a common factor c-1. That root corresponds to y=x, which makes the original equation indeterminate. Otherwise: (c-1) = c2(c+1) OTOH, where y < 0: dy/dx = -(√((y-x)2+y2)+y)/(y-x) Put y = cx: c = -(√((c-1)2+c2)+c)/(c-1) c(c-1) - c = -√((c-1)2+c2) = c2 -2c etc. |
| Jul19-12, 06:00 PM | #5 |
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c+x*dc/dx =(√((c-1)2+c2)-c)/(c-1) x*dc/dx=(√((c-1)2+c2)-c^2)/(c-1) Then, I am quite at a loss trying to integrate and solve that for c |
| Jul19-12, 06:27 PM | #6 |
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Recognitions:
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| nonlinear, ode, radical |
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