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2 years ago

Solve the following system of equations by linear combination: 5x 2y = 14 y = x – 7

1 answer:

4 0

I would use subsitution
y=x-7
sub x-7 for every x
5x+2(x-7)=14
5x+2x-14=14
add 14 to both sides
7x=28
divide both sides by 7
x=4
sub
y=x-7
y=4-7
y=-3

x=4
y=-3
(4,-3)

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Power series of y''+x^2y'-xy=0

Ray Of Light [21]

Assuming we're looking for a power series solution centered around $x=0$ , take

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}$
$y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}$

Substituting into the ODE yields

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0$

The first series starts with a constant term; the second series starts at $x^2$ ; the last starts at $x^1$ . So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a $x^2$ term. We have

$\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}$

$\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}$

Re-index the first sum to have it start at $n=1$ (to match the the other two sums):

$\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0$

Consolidate into one series starting $n=1$ :

$\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0$

Suppose we're given initial conditions $y(0)=a_0$ and $y'(0)=a_1$ (which follow from setting $x=0$ in the power series representations for $y$ and $y'$ , respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when $n=3k-2$ , i.e. $n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0$

and so on, so that $a_{3k-2}=0$ except for when $k=1$ .

Now let's consider $n=3k-1$ , or $n\in\{2,5,8,11,\ldots\}$ . We know that $a_2=0$ , and from the recurrence it follows that $a_{3k-1}=0$ for all $k$ .

Finally, take $n=3k$ , or $n\in\{0,3,6,9,\ldots\}$ . We have a solution for $a_3$ in terms of $a_0$ , so the next few terms ( $k=2,3,4$ ) according to the recurrence would be

$a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}$
$a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}$
$a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for $n=3k$ as

$a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}$
$a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

So the series solution to the ODE is given by

$y=\displaystyle\sum_{n\ge0}a_nx^n$
$y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at $a_0=y(0)=1$ and $a_1=y'(0)=2$ overlaid with the series solution (orange) with $n=3$ and $n=6$ . (Note the rapid convergence.)