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spaghetti3451
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Homework Statement
Solve the following differential equation by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution.
y'' = - 4y
Homework Equations
The Attempt at a Solution
[itex]
y'' = - 4y \\
\frac{d^{2}y}{dx^{2}} + 4y = 0 \\
Characteristic\ equation : λ^{2} + 4 = 0 \Rightarrow λ = ± 2i \\
y = A\cos(2x) + B\sin(2x) \\
y = A\ \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m}}{(2m)!} + B\ \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m+1}}{(2m+1)!}
[/itex][itex]
Let\ y\ = \sum^{n = ∞}_{n = 0} a_{n}x^{n} \\
\sum^{n = ∞}_{n = 0} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 0} a_{n}x^{n} = 0 \\
\sum^{n = ∞}_{n = 2} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 0} a_{n}x^{n} = 0 \\
\sum^{n = ∞}_{n = 2} n(n-1)a_{n}x^{n-2} + 4\sum^{n = ∞}_{n = 2} a_{n-2}x^{n-2} = 0 \\
\sum^{n = ∞}_{n = 2} [ n(n-1)a_{n} + 4 a_{n-2} ] x^{n-2} = 0 \\
n(n-1)a_{n} + 4 a_{n-2} = 0,\ n = 2, 3, 4, ... \\
a_{n} = - \frac{(2)^{2}}{n(n-1)} a_{n-2},\ n = 2, 3, 4, ... \\
a_{2m} = - \frac{(2)^{2}}{(2m)(2m-1)} a_{2m-2}\ and\ a_{2m+1} = - \frac{(2)^{2}}{(2m+1)(2m)} a_{2m-1},\ m = 1, 2, 3, ... \\
a_{2m} = (-1)^{m} \frac{(2)^{2m}}{(2m)!} a_{0}\ and\ a_{2m+1} = (-1)^{m} \frac{(2)^{2m}}{(2m+1)!} a_{1},\ m = 1, 2, 3, ... \\
y = a_{0} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2x)^{2m}}{(2m)!} + a_{1} \sum^{m = ∞}_{m = 0} (-1)^{m} \frac{(2)^{2m}(x)^{2m+1}}{(2m+1)!} \\
[/itex]
So, the solutions don't match because of the missing factor of 2 in the second summation of the series solution.
Could you please help me out?
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