# Find Boundary value for nonlinear differential equation?

Posted 1 year ago
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 y(0)=2; y'(x)=(0.004-(0.022^220^2)/(y(x)^(10/3)))/(1-((20^2)/(9.81(y(x))^3))), y(L)=3I wanna compute value L.When I enter above equation, it just gives another differential equation.How can I compute L with Wolfram Alpha? Answer
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Posted 1 year ago
 Do you what compute this: 0.022^220^2 value ? It's: 1.810305182819054738210^-80227 Answer
Posted 1 year ago
 It seems there is no solution: lsg = NDSolve[ {y'[x] == (0.004 - (0.022^220^2)/(y[x]^(10/3)))/(1 - ((20^2)/(9.81 (y[x])^3))), y == 2}, y, {x, 0, 5000}]; fy = y /. lsg[[1, 1]]; Plot[fy[x], {x, 0, 5000}] Look at your y'. Map[Together, (0.004 - (0.022^220^2)/(y[ x]^(10/3)))/(1 - ((20^2)/(9.81 (y[x])^3))), {1}] // Chop // FullSimplify You start with y == 2. So y' is negative and y will never reach values to change the sign of y', so y will become smaller and smaller. Answer
Posted 1 year ago
 Here is another approach to your problem (please name the plot given after NDSolve as pl1) e1 = (0.004 - (0.022^220^2)/(y^(10/3)))/(1 - ((20^2)/(9.81 (y)^3))); e2 = e1 // Chop // FullSimplify SetOptions[Integrate, GenerateConditions -> False]; jj = Integrate[(y^3 - b)/(a y^3), {y, 2, yy}] lsg = Solve[jj == t, yy][[3, 1]]; fy = yy /. lsg; fy /. t -> 0 /. a -> .004 /. b -> 40.7747 Chop[fy /. t -> 0 /. a -> .004 /. b -> 40.7747] D[fy, t] - a fy^3/(fy^3 - b) // FullSimplify ffy = fy /. a -> .004 /. b -> 40.7747; pl2 = Plot[Chop[ffy /. t -> tt], {tt, 300, 2700}, PlotStyle -> {Thick, Red}] You see that the (appoximated) diff. equation is fulfilled and the solution is essentially equal to the result of NDSolve: Show[pl1, pl2] y[L] =3 is reached for a negative argument: FindRoot[ffy == 3, {t, -473}] (* {t->-457.8940972222226-4.814826364873225*^-13 I} *) Answer