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# how to enter y"+y(a-b(sqrt(c-y^2)))=0 into wolf. alpha and get a solution?

Posted 9 years ago
 I tried y"+y(a-b(sqrt(c-y^2)))=0 y"+y(a-b(c-y^2)^.5)=0 y"+y(a-b[sqrt(c-y^2)])=0 etc., But Wolfram Alpha wouldn't solve the equation and, for many of the attempts, interpreted the input as an equation in y(c). What am I doing wrong? Thanks in advance. Edit: entering y"+(a-(sqrt(c+y^2)))y=0 caused wolfram to at least identify it as a diff. eqn, but it still interpreted it as an eqn. in y(c), and couldn't solve it 2nd edit: entering y"+(1-(sqrt(1+y^2)))y=0 allowed wolfram to attempt a solution as an eqn. of y(x), but no closed form solution, just graphs... perhaps there is no closed form solution for this eqn.? 3rd edit: y"+(1-(sqrt(1+y^2)))y=0,y(0)=1,y'(0)=0 allowed wolfram to attempt a solution as an eqn. of y(x), but it is still calculating (after 6min)... perhaps wolfram can't solve this? 4th edit: This isn't an edit, it's a response to David Reiss. I was unable to post a normal response in the forum, don't know why. Thanks for the help, but I've never used Mathematica, so I'm afraid I don't understand most of your response. I inputed y(x)"+(a-b(sqrt(c+(y(x)^2))))y(x)=0,y(0)=1,y'(0)=0 which allowed me to input the proper eqn. (with the constants), and the program interpreted the input correctly as an eqn. in y(x), but it still can't solve it...just keeps thinking. I would like to understand your response. Perhaps if you explained the details of what is meant by the notation. But since I don't understand Mathematica, this might be difficult. I do appreciate your response.
 To get W|A to recognize things properly, give your y functions explicit arguments as in y(x)"+(a-(sqrt(c+y(x)^2)))y(x)=0 or y[x]"+(a-(sqrt(c+y[x]^2)))y[x]=0 it may well be that your differential equation does not have an explicit closed form solution. In Mathematica trying DSolve[y''[x] + (1 - Sqrt[1 + y[x]^2]) y[x] == 0, y[x], x] grinds along for some while--it hasn't yet spit out an answer after several minutes. The first order version of this (i.e., only a single derivative for the first term), DSolve[y[x] + (1 - Sqrt[1 + y[x]^2]) y[x] == 0, y[x], x] which is trivially integrable yields {{y[x] -> InverseFunction[-(( 1 - Log[#1] #1^2 + Log[1 + Sqrt[1 + #1^2]] #1^2 + Sqrt[ 1 + #1^2])/(2 #1^2)) &][x + C]}} because the integration cannot be analytically inverted to get y in terms of x. Ok, now the second order problem yielded a solution: here it is: Solve[Integrate[1/Sqrt[C + 2*((-(1/2))*K^2 + (1/3)*(1 + K^2)^(3/2))], {K, 1, y[x]}]^2 == (x + C)^2, y[x]] I.e. it is implicit and in terms of inverting a function expressed as an integral that Mathematica does not know how to do. I.e., Integrate[1/Sqrt[c + 2*(-x^2/2 + (1 + x^2)^(3/2)/3)], x]