- #1
Bucky
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Homework Statement
Use the integrating factor to solve the first order differential equation
dy/dx + (2/x)y = 1 - 2/x
subject to the initial condition y(1) = 2
The Attempt at a Solution
I have the full answer (it's an example), but my lecturer is notorious for omitting steps, to my constant irritation. It might be fine for pure mathmaticians, but for some of us comp sci fakers, it makes everything much more difficult
[itex]
P = \frac{2}{x} [/itex]
[itex]
Q = 1-\frac{2}{x}? or just \frac{2}{x} ?
[/itex]
Integrating factor is therefore
[itex]F = e^{2 \int \frac{1}{x}}
...
F = x^2
[/itex]
Upon multiplying by [itex]F = x^2 [/itex], we find
[itex] x^2 \frac{dy}{dx} + 2xy= x^2 - 2x [/itex]
Where does the 2xy go?
[itex]
((x^2) y)' = x^2 + 2x
[/itex]
so that
[itex]
x^2 y = \int (x^2 - 2x) dx [/itex]
[itex]
x^2y = \frac{x^3}{3} - x^2 + C
[/itex]
when x = 1, y = 2
[itex]
x^2 y = \frac{x^3}{3} - x^2 + C
2 = \frac{1}{3} -1 + C
C = \frac{8}{3}
[/itex]
therefore, sollution is
[itex]
y = \frac{x}{3} - x + \frac {8}{3x^2}
[/itex]
where did the x^2 come from at the end?
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