Mathematics, 16.10.2019 06:00 nina4552

Find the first three nonzero terms in each of two solutions (noy1t multiples of each other) xy''+y'-y=0

i know y1 and a_n(r) but i don't know how to differntiate a_n(r).

as i know, a_n(r)= 1/(r++n)

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It looks like you're trying to solve the ODE via Frobenius' method. It's not immediately clear where the solution is to be centered, but I think we can safely assume it's to be taken about x=0

. Dividing through by

, we get

$y''+\dfrac{y'}x-\dfrac yx=0$

and we observe that x=0

is a regular singular point.

Now, taking a solution of the form

$y=\displaystyle\sum_{n\ge0}a_nx^{n+r}$
$\implies y'=\displaystyle\sum_{n\ge0}(n+r)a_nx^{n+r-1}$ $\implies y''=\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-2}$

and substituting into the ODE gives

$\displaystyle\sum_{n\ge0}(n+r)(n+r-1)a_nx^{n+r-1}+\sum_{n\ge0}(n+r)a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle\sum_{n\ge0}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge0}a_nx^{n+r}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}(n+r)^2a_nx^{n+r-1}-\sum_{n\ge1}a_{n-1}x^{n+r-1}=0$
$\displaystyle r^2a_0x^{r-1}+\sum_{n\ge1}\bigg((n+r)^2a_n-a_{n-1}\bigg)x^{n+r-1}=0$

The indicial polynomial admits only one root (of multiplicity two) at r=0

, so in fact a regular power series solution will exist. So in fact the ODE above reduces to

$\displaystyle\sum_{n\ge1}\bigg(n^2a_n-a_{n-1}\bigg)x^{n-1}=\sum_{n\ge0}\bigg((n+1)^2a_{n+1}-a_n\bigg)x^n=0$

leaving us with the recurrence

$a_{n+1}=\dfrac{a_n}{(n+1)^2}$

We can solve this by recursively substituting the right hand side:

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

and so one of the solutions will be

$y_1(x)=\displaystyle a_0\sum_{n\ge0}\frac{x^n}{(n!)^2}$

Now the literature on the Frobenius method when the indicial polynomial has repeated roots suggests a second solution will take the form

$y_2(x)=Cy_1\ln x+\displaystyle\sum_{n\ge0}b_nx^n$

but it's not clear to me *why* this works.

Answer from: Quest

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation:

Which are the possible side lengths of a triangle?

Answer from: Quest

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

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