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Find the general solution?

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1 Answer

Use the integrating factor method.

μ dy/dx + 3μy = μ e3x

Let μ satisfy dμ/dx = 3μ.

Substitution yields μ dy/dx + y dμ/dx = μe3x.

The left side is a product rule for (d/dx) μy so you can integrate both sides.

μy = ∫ μe3x dx

y = (1/μ) ∫ μe3x dx

Now find μ from it's differential equation which is separable.

dμ/dx = 3μ

dμ/μ = 3 dx.

∫ dμ/μ = ∫ 3 dx

ln |μ| = 3x + C

μ = e3x+C = ece3x = Ae3x

Plug in any specific μ, say μ = e3x into the original solution of y.

y = e-3x ∫ e3xe3x dx = e-3x ∫ e6xdx = e-3x (e6x/6 + C) = e3x/6 + Ce-3x

Check:

(d/dx) (e3x/6 + Ce-3x) + 3(e3x/6 + Ce-3x) = e3x/2 - 3Ce-3x + e3x/2 + 3Ce-3x = e3x.