## Second order, nonlinear differential equation help

I've been trying to solve this one for a while. My professor wasn't even sure how to do it. Any suggestions?
y''+t*y+y*y'=sin(t)

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 Quote by cpyles1 I've been trying to solve this one for a while. My professor wasn't even sure how to do it. Any suggestions? y''+t*y+y*y'=sin(t)
Yeah. How about a power series solution? Huh? What do you mean that yy' thing? No? How about the Mathematica code below then. Looks pretty close to me.

Code:
Remove[a];

nmax = 25;

myleftside = Sum[n*(n - 1)*Subscript[a, n]*t^(n - 2), {n, 0, nmax}] +
Sum[Subscript[a, n - 3]*t^(n - 2), {n, 3, nmax + 2}] +
Sum[Subscript[a, k]*Subscript[a, n - k]*(n - k)*t^(n - 1), {n, 0, nmax + 1}, {k, 0, n}];

myrightside = Sum[((-1)^n*t^(2*n + 1))/(2*n + 1)!, {n, 0, nmax}];

myclist = Flatten[Table[Solve[Coefficient[myleftside, t, n] == Coefficient[myrightside, t, n],
Subscript[a, n + 2]], {n, 0, nmax}]];

Subscript[a, 0] = 0;
Subscript[a, 1] = 1;

mysec = Table[Subscript[a, n] = Subscript[a, n] /. myclist, {n, 2, nmax}];

thef[t_] := Sum[Subscript[a, n]*t^n, {n, 0, nmax}];

p1 = Plot[thef[t], {t, 0, 2}, PlotStyle -> Red];

mysol = NDSolve[{Derivative[2][y][t] + t*y[t] + y[t]*Derivative[1][y][t] == Sin[t], y[0] == 0,
Derivative[1][y][0] == 1}, y, {t, 0, 2}];

p2 = Plot[y[t] /. mysol, {t, 0, 2}, PlotStyle -> Blue];

Show[{p1, p2}]

 Tags nonlinear, second order