hF(a,n) = \sum_{i=0}^{n}a^i = 1 + a \sum_{i=0}^{n-1}a^i = 1 + a F(a,n-1)
F(a,2k+1) = \sum_{i=0}^{2k+1}a^i = \sum_{i=0}^{k}a^{2i} + \sum_{i=0}^{k}a^{2i+1} = \sum_{i=0}^{k}a^{2i} + a \sum_{i=0}^{k}a^{2i} = (1 + a) \sum_{i=0}^{k}a^{2i} = (1 + a) F(a^2,k)
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x^{^{n}} =\left\{\begin{matrix}
{{(x^2)}^k} &, n = 2k \\
x\cdot x^{n-1} &, n = 2k+1
\end{matrix}\right.
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(\frac{1}{x} + \frac{1}{x+5})*6 = 1
\frac{x+5 + x}{x(x+5)}*6 = 1
\frac{2x+5}{x^{2}+5x}*6 = 1
12x + 30 = x^{2} + 5x
x^{2} - 7x - 30 = 0
(x-10)(x+3)= 0
x_{1} = 10;x_{2} = -3;
\[y'' - \left(2+c\right)y' + \left(1+c\right)y = e^{\left(1+2c\right)x}\]
for $c \neq 0$.
Solve the auxiliary equation:
\[\lambda^2 - \left(2+c\right)\lambda + \left(1+c\right) = 0\]
\[\lambda = \frac{2+c \pm \sqrt{\left(2+c\right)^2 - 4\left(1+c\right)}}{2}\]
\[\lambda = \frac{2 + c \pm c}{2}\]
Hence the homogeneous solution is of the form
\[Ae^x + Be^{\left(1+c\right)x}\]
To find the particular integral of $e^{\left(1+2c\right)x}$, try trial solution $y_p = Ce^{\left(1+2c\right)x}$:
\[\left(1+2c\right)^2Ce^{\left(1+2c\right)x} - \left(2+c\right)\left(1+2c\right)Ce^{\left(1+2c\right)x} + \left(1+c\right)Ce^{\left(1+2c\right)x} = e^{\left(1+2c\right)x}\]
\[(4c^2 + 4c + 1 - 4c - 2 - c - 2c^2 + 1 + c)Ce^{\left(1+2c\right)x} = e^{\left(1+2c\right)x}\]
\[C = \frac{1}{2c^2}\]
We can now deduce that
\[\boxed{y = Ae^x + Be^{\left(1+c\right)x} + \frac{1}{2c^2}e^{\left(1+2c\right)x}}\]
a) Show that there is a solution of the form $y=f(x,c)$ where
\[f(x,c) = Ae^x + B\frac{e^x}{c}\left(e^{cx}-1\right) + \frac{e^x}{2c^2}\left(e^{2cx} - 2e^{cx} + 1\right)\]
for any $A$, $B$ and $c$ ($c \neq 0$).
By expanding we get
\[f(x,c) = Ae^x + \frac{B}{c}e^{\left(1+c\right)x} - B\frac{e^x}{c} + \frac{1}{2c^2}e^{\left(1+2c\right)x} - \frac{1}{c^2}e^{\left(1+c\right)x} + \frac{1}{2c^2}e^x\]
\[\boxed{f(x,c) = \left(A - \frac{B}{c} + \frac{1}{2c^2}\right)e^x + \left(\frac{B}{c} - \frac{1}{c^2}\right)e^{\left(1+c\right)x} + \frac{1}{2c^2}e^{\left(1+2c\right)x}}\]
which is consistent with the general solution derived earlier.
b) Find the limit of $f(x,c)$ as $c \rightarrow 0$. Hence or otherwise find the complementary function and particular integral in the case $c=0$.
\[\lim_{c \to 0} f(x,c) = Ae^x + Be^x \lim_{c \to 0} \frac{e^{cx} - 1}{c} + \frac{e^x}{2} \lim_{c \to 0} \frac{e^{2cx}-2e^{cx}+1}{c^2}\]
\[\lim_{c \to 0} f(x,c) = Ae^x + Bxe^x + \frac{1}{2}x^2e^x\]
Hence, when $c = 0$,
\[\boxed{y = Ae^x + Bxe^x + \frac{1}{2}x^2e^x}\]
