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Solve $y'' -2*(x^4-e)y = 0$ 2nd order non-linear differential equation?

GROUPS:

I want to solve the below differential equation:

$$y'' -2*(x^4-e)y = 0$$

with i.c. y(0) = 1 , y'(0) = 0

the values of e can be different, and I want to plot the result for different given values of e. I would be grateful if someone could help me answer this question.

Answer
11 months ago

ParametricNDSolve is the function that you are looking for. Use the second example on the page:

sol = ParametricNDSolve[{y''[t] - 2*(t^4-a)  y[t] == 0, y[0] == 1, y'[0] == 0},y, {t, 0, 1}, {a}]
Plot[Evaluate[Table[y[a][t] /. sol, {a, -1, 1, .1}]], {t, 0, 1}, PlotRange -> All]
POSTED BY: Shenghui Yang
Answer
11 months ago

thank you! do you also know if i can give a vector of values to a ?

Answer
11 months ago
sol = DSolve[{y''[t] - 2*(t^4 - e) y[t] == 0, y[0] == 1, y'[0] == 0}, y, t]
sol = First@sol;

With[{erange = Range[2, -2, -1]},
 Plot[
  Evaluate@Table[
    y[t] /. sol,
    {e, erange}],
  {t, 0, 2},
  PlotLegends -> (HoldForm[e = #] & /@ erange)
  ]
 ]

plot of solutions

Note the result of DSolve is a DifferentialRoot, but it can be evaluated like other expressions.

POSTED BY: Michael Rogers
Answer
11 months ago

This is exactly what I want except that it doesn't work with non-integer values for 'e' :( Can you help me to fix that? Id really appreciate it

Answer
11 months ago

You can use Rationalize[.., 0] or SetPrecision[.., Infinity] to convert approximate numbers to the exact numbers required by DifferentialRoot[]. (There is a slight difference between the two functions, on the order of 10^-16 or less, but it makes no difference in plotting these solutions.)

SeedRandom[2];
With[{erange = Sort[RandomReal[{-3, 3}, 6]]},
 Plot[
  Evaluate@Table[
    y[t] /. sol,
    {e, Rationalize[erange, 0]}],
  {t, 0, 2},
  PlotLegends -> (HoldForm[e = #] & /@ erange
    )
  ]
 ]

enter image description here

POSTED BY: Michael Rogers
Answer
11 months ago

Thank you so much this was exactly what i needed!

Answer
11 months ago

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