NewtonianAlch
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Homework Statement
Solve [itex]\frac{dy}{dx}[/itex] - 2y = x[itex]^{2}[/itex]e[itex]^{2x}[/itex]
The Attempt at a Solution
Integrating factor = e[itex]^{2x}[/itex]
So we multiply through the given equation by the integrating factor and get:
e[itex]^{2x}[/itex][itex]\frac{dy}{dx}[/itex] - 2e[itex]^{2x}[/itex]y = x[itex]^{2}[/itex]e[itex]^{4x}[/itex]
Contract the left-hand side via the chain rule to get:
[itex]\frac{d}{dx}[/itex](e[itex]^{2x}[/itex]y) = x[itex]^{2}[/itex]e[itex]^{4x}[/itex]
Integrate both sides
e[itex]^{2x}[/itex]y = [itex]\frac{1}{32}e^{4x}[/itex](8x[itex]^{2}[/itex]-4x+1)+C
Now divide through by e[itex]^{2x}[/itex] and the equation definitely does not equal what Wolfram Alpha gives as the solution:
y = [itex]\frac{1}{3}[/itex]e[itex]^{2x}[/itex]x[itex]^{3}[/itex]+Ce[itex]^{2x}[/itex]
I checked some of the parts individually with Wolfram, such as the integration of the right-hand side and that was correct, so I'm not too sure what's causing the difference in answers.