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mariarad [96]
3 years ago
5

Power Series Differential equation

Mathematics
1 answer:
KatRina [158]3 years ago
4 0
The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for y

\displaystyle\sum_{n\ge2}\bigg((n-3)(n-2)a_n+(n+3)(n+2)a_{n+3}\bigg)x^{n+1}+2a_2+(6a_0-6a_3)x+(6a_1-12a_4)x^2=0

which indeed gives the recurrence you found,

a_{n+3}=-\dfrac{n-3}{n+3}a_n

but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that a_2=0, and substituting this into the recurrence, you find that a_2=a_5=a_8=\cdots=a_{3k-1}=0 for all k\ge1.

Next, the linear term tells you that 6a_0+6a_3=0, or a_3=a_0.

Now, if a_0 is the first term in the sequence, then by the recurrence you have

a_3=a_0
a_6=-\dfrac{3-3}{3+3}a_3=0
a_9=-\dfrac{6-3}{6+3}a_6=0

and so on, such that a_{3k}=0 for all k\ge2.

Finally, the quadratic term gives 6a_1-12a_4=0, or a_4=\dfrac12a_1. Then by the recurrence,

a_4=\dfrac12a_1
a_7=-\dfrac{4-3}{4+3}a_4=\dfrac{(-1)^1}2\dfrac17a_1
a_{10}=-\dfrac{7-3}{7+3}a_7=\dfrac{(-1)^2}2\dfrac4{10\times7}a_1
a_{13}=-\dfrac{10-3}{10+3}a_{10}=\dfrac{(-1)^3}2\dfrac{7\times4}{13\times10\times7}a_1

and so on, such that

a_{3k-2}=\dfrac{a_1}2\displaystyle\prod_{i=1}^{k-2}(-1)^{2i-1}\frac{3i-2}{3i+4}

for all k\ge2.

Now, the solution was proposed to be

y=\displaystyle\sum_{n\ge0}a_nx^n

so the general solution would be

y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+a_6x^6+\cdots
y=a_0(1+x^3)+a_1\left(x+\dfrac12x^4-\dfrac1{14}x^7+\cdots\right)
y=a_0(1+x^3)+a_1\displaystyle\left(x+\sum_{n=2}^\infty\left(\prod_{i=1}^{n-2}(-1)^{2i-1}\frac{3i-2}{3i+4}\right)x^{3n-2}\right)
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3x - 2(2x - 5) = 2(x + 3) - 8
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<h3>Solve for x</h3>
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3 0
2 years ago
Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x. (
True [87]
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.

So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))

now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16 

Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84

So we have our linear approximation for the two. 

If you wanted to, you could check your answer by finding g(x).  Since you know g'(x), take the antiderivative and we will get 
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35

So just to check our linear approximations use that to find g(2.99) and g(3.01)

g(2.99) = -5.1597
g(3.01) = -4.8397

So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer.  Not a bad method if you ever need to use it. 
5 0
3 years ago
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