Question

Differential Equation: x^2 y' + y = x^2 e^(-x)
What I'v done:
y' + 1/x^2 * y = e^(-x)
Integrating factor: e^(-1/x)
Which leads to: Integral of: e^(-x) * e^(-1/x)
Wolfram says that there are no elementary function for this integral. Solving the homogeneous eqn:
y' + 1/x^2 y = 0
Gives y_h = Ce^(1/x) which is correct as far as I know. But how can I find the particular solution?

Answer 1

wolframalpha is right

Answer 2

Your y_h solution is right.
The particular solution is given by a non-elementary integral
\[
y_p=e^{\frac{1}{x}} \int_1^x
e^{-x-\frac{1}{x}} \, dx
\]

Answer 3

\[
y_p=e^{\frac{1}{x}} \int_1^x
e^{-t-\frac{1}{t}} \, dt
\]

Answer 4

May be you should try Series solution of it

Answer 5

Here a series solution up to power of 10
\[y(x)=x^2-3 x^3+\frac{19 x^4}{2}-\frac{229
x^5}{6}+\frac{1527
x^6}{8}-\frac{137431
x^7}{120}+\\ \frac{5772103
x^8}{720}-\frac{107745923
x^9}{1680}+\\ \frac{23273119369
x^{10}}{40320}+O\left(x^{11}\right)\]

Answer 6

The above solution was generated by mathematica using the script below


n = 10;
y[x_] = Sum[a[i] x^i, {i, 0, n}] + O[x]^(n + 1);
eqns = LogicalExpand[x^2 y'[x] + y[x] == x^2 E^-x];
ans = Solve[eqns];
g[x_] = y[x] /. ans[[1]]

Answer 7

Thank you! But there exists a simple solution:
Ce^(1/x) + x*e^(1/x)
=> y_p = x * e^(1/x)
y_p satisfies the equation but I can't figure out how the get it!