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3 years ago

Can someone please help me with 2 and explain the steps to 3 thanks!

Mathematics

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

#2 is a ratio so you gotta add.

#3 it’s just adding it for example 2p + 5a = 7pa

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$y=\displaystyle\sum_{n=0}^\infty a_nx^n$

then

$y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n$

The ODE in terms of these series is

$\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0$

$\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}$

We can solve the recurrence exactly by substitution:

$a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0$

$\implies a_n=\dfrac{(-2)^n}{n!}a_0$

So the ODE has solution

$y(x)=\displaystyle a_0\sum_{n=0}^\infty\frac{(-2x)^n}{n!}$

which you may recognize as the power series of the exponential function. Then

$\boxed{y(x)=a_0e^{-2x}}$

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