How to find the general solution of xy'-4y=e^x using the associated homogeneous equation ?

Question

so far I've done:

xy'-4y=e^x

xy_H-4y_H=0

ln|y_H|=4lnx+c    x⁴

|y_H|=Ae^{(x^4)} 

y(x)=A(x)e^{(x^4)}

 

y'=A'e(x^4)+4x³A(x)e^{(x^4)}=A'e^{(x^4)}+4x³y

x(A'e^{(x^4)}+4x³y)-4A(x)e^{(x^4)}=e^x

Answer 1

First gain familiarity with the actual solution.

From xy' − 4y = e^x, go to y' − (4/x)y = e^x/x, which is in the form of a linear first-order equation or

dy/dx + P(x)y = Q(x).

For y' − (4/x)y = e^x/x, obtain an integrating factor by writing e^∫(-4/x)dx equal to e^{-4ln |x|} or |x|^-4 or 1/x⁴.

Now build the "primitive" solution as y(1/x⁴) = ∫(1/x⁴)(e^x/x)dx + C₁; multiply through this last by x⁴ and write y = x⁴∫(1/x⁴)(e^x/x)dx + C₁x⁴.

Then isolate ∫(1/x⁴)(e^x/x)dx. A tremendous integration by parts expresses the solution to this integral as Ei(x)/24 − (x³ + x² + 2x + 6)e^x/24x⁴ + C₂.

{Note that Ei(x) above stands for "Exponential Integral with respect to x" and is equal to ∫(e^x/x)dx which translates via Maclaurin's Series to ln |x| + x¹/1(1!) + x²/2(2!) + x³/3(3!) + ... + C₃.}

Now develop y = x⁴∫(1/x⁴)(e^x/x)dx + C₁x⁴ as: y = x⁴{Ei(x)/24 − (x³ + x² + 2x + 6)e^x/24x⁴ + C₂} +C₁x⁴.

Next, y = x⁴Ei(x)/24 − [(x³ + x² + 2x + 6)e^x/24 +C₂x⁴] + C₁x⁴.

Then, y = x⁴Ei(x)/24 − x³e^x/24 − x²e^x/24 − 2xe^x/24 − 6e^x/24 + (C₁ + C₂)x⁴.

Final simplification gives y = Cx⁴ + x⁴Ei(x)/24 − x³e^x/24 − x²e^x/24 − xe^x/12 − e^x/4.