1
$\begingroup$
DSolve[y''[x] + (a - (b*x^2 - c)*x^2)*y[x] == 0, y[x], x]

I couldn't solve this equation, please help

$\endgroup$
  • 1
    $\begingroup$ There is no solution for the anharmonic oscillator in terms of the elementary or any standard special function as far as I know. $\endgroup$ – Nikolay Gromov Sep 17 '15 at 12:13
  • 1
    $\begingroup$ I get a solution in terms of DifferentialRoot. What version are using? $\endgroup$ – Michael E2 Sep 17 '15 at 13:31
2
$\begingroup$

Mathematica can't solve analytically,but numerical yes can.

Clear["Global`*"]
a = 3;
b = 2;
c = 1;
eq = {y''[x] + (a - (b*x^2 - c)*x^2)*y[x] == 0, y[1] == 0, y'[0] == 1};
sol = First@NDSolve[eq, y, {x, -2, 2}];
Plot[Evaluate[y[x] /. sol], {x, -2, 2}, PlotRange -> All, 
PlotLegends -> {"y(x)"}, AxesLabel -> {x, y[x]}]

enter image description here

MAPLE analytical solution:

enter image description here

y[x] == 
C[1]*HT[(2^(1/3)*3^(2/3)*(4*a*b + c^2))/(8*b^(4/3)), 
0, (-c*2^(1/3)*3^(1/3))/(2*b^(2/3)), (-2^(1/3)*3^(2/3)*b^(1/6)*x)/
3]*Exp[(x*(2*b*x^2 - 3*c))/(6*Sqrt[b])] + 
C[2]*HT[(2^(1/3)*3^(2/3)*(4*a*b + c^2))/(8*b^(4/3)), 
0, (-c*2^(1/3)*3^(1/3))/(2*b^(2/3)), (2^(1/3)*3^(2/3)*b^(1/6)*x)/
3]*Exp[(-(b*x^2 - 3/2*c)*x)/(3*Sqrt[b])]

You can be expressed this equation by the series:

$$y(x)=c_1 \text{HT1} \exp \left(\frac{x \left(2 b x^2-3 c\right)}{6 \sqrt{b}}\right)+c_2 \text{HT2} \exp \left(-\frac{\left(b x^2-\frac{3 c}{2}\right) x}{3 \sqrt{b}}\right)$$

were HT1 and HT2 is:

enter image description here

and

enter image description here

References for Triconfluent Heun function:

1.Decarreau, A.; Dumont-Lepage, M.C.; Maroni, P.; Robert, A.; and Ronveaux, A. "Formes Canoniques de Equations confluentes de l'equation de Heun". Annales de la Societe Scientifique de Bruxelles. Vol. 92 I-II, (1978): 53-78.

2.Ronveaux, A. ed. Heun's Differential Equations. Oxford University Press, 1995.

3.Slavyanov, S.Y., and Lay, W. Special Functions, A Unified Theory Based on Singularities. Oxford Mathematical Monographs, 2000.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.