Question

solve by using power series method: 2y'-y=sinh(x). compute the first 6 coefficients (a0-a5)

Answer 1

$$2y'-y=\sinh x$$Recall that we know \(\sinh x=\frac12(e^x-e^{-x})\). To determine a power series representation, first consider expansions for \(e^x,e^{-x}\):$$e^x=\sum_{n=0}^\infty\frac1{n!}x^n\\e^{-x}=\sum_{n=0}^\infty\frac1{n!}(-x)^n=\sum_{n=0}^\infty\frac{(-1)^n}{n!}x^n$$Now consider their difference:$$e^x-e^{-x}=\sum_{n=0}^\infty\frac1n\left(x^n-(-1)^nx^n \right)$$We know that for even \(n\) we have \((-1)^n=1\) so \(x^n=(-1)^nx^n=x^n-x^n=0\). For odd \(n\) we have \((-1)^n=-1\) so we find \(x^n-(-1)^nx^n=x^n+x^n=2x^n\). We can then rewrite our series to consider only odd terms:$$e^x-e^{-x}=\sum_{n=0}^\infty\frac2{(2n+1)!}x^{2n+1}$$Since \(\sinh x\) is exactly \(1/2\) of that, we find:$$\sinh x=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{2n+1}$$

Answer 2

So thus far we know:$$2y'-y=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{2n+1}$$

Answer 3

Presume we have an analytic solution \(y\) so we may express it as a power series \(y=\sum\limits_{n=0}^\infty a_nx^n\). It follows that our derivative ought to just be \(y'=\sum\limits_{n=0}^\infty(n+1)a_{n+1}x^n\)

Answer 4

Substitute into our equation to find:$$2\sum_{n=0}^\infty(n+1)a_{n+1}x^n-\sum_{n=0}^\infty a_nx^n=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{2n+1}\\\sum_{n=0}^\infty(2(n+1)a_{n+1}-a_n)x^n=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{2n+1}$$Because each power of \(x\) is linearly independent of the others we can safely equate coefficients, yielding a linear recurrence relation in \(a_n\):$$2(n+1)a_{n+1}-a_n=\frac1{(2n+1)!}\\2na_{n+1}+2a_{n+1}-a_n=\frac1{(2n+1)!}$$

Answer 5

this helped a lot. Thank you very much

Answer 6

no problem :-)

Answer 7

@goalie2012 wait I think I did the coefficients wrong... I matched the wrong powers

Answer 8

@goalie2012 the right has no even powers of \(x\) hence our left-hand coefficients must be \(0\) for \(n=0,2,4\)

Answer 9

@oldrin.bataku so I only need to calculate 0, 2, and 4, or do you think I need to do 0,2,4,6,8, and 10?

Answer 10

@goalie2012 you need to compute them all but the equations for \(n=0,2,4\) should be \(2(n+1)a_{n+1}-a_n=0\) whereas our equations for \(n=1,3,5\) should be \(2(n+1)a_{n+1}-a_n=1/n!\) -- sorry!!!!

Answer 11

not \[1/(2n+1)!\]?

Answer 12

@oldrin.bataku not
1/(2n+1)!
?

Answer 13

@goalie2012 nope; the \(1/(2n+1)!\) is for the \(n\) on our right-hand side, which gives the coefficient of \(x^{2n+1}\)

Answer 14

on the left-hand side, however, we have the coefficients of \(x^n\)...

Answer 15

$$\sum_{n=0}^\infty(2(n+1)a_{n+1}-a_n)x^n=\sum_{n=0}^\infty\frac1{(2n+1)!}x^{2n+1}$$So we see our right-hand side lacks even powers of \(x\), implying our left-hand expression for coefficients is \(0\) for \(n=0,2,4,\dots\):$$2(n+1)a_{n+1}-a_n=0\\2a_1-a_0=0\\6a_3-a_2=0\\10a_3-a_4=0$$For *odd* powers of \(x\) observe the coefficient on the right is merely \(1/n!\):$$2(n+1)a_{n+1}-a_n=\frac1{n!}$$So, we let \(n=1,3,\dots\):$$4a_2-a_1=\frac1{1!}=1\\8a_4-a_3=\frac1{3!}=\frac16$$

Answer 16

$$2a_1-a_0=0\implies a_1=\frac{1+a_0}2\\4a_2-a_1=1\implies a_1=\frac{1+a_1}4=\frac{2+1+a_0}8=\frac{3+a_0}8\\6a_3-a_2=0\implies a_3=\frac{a_2}6=\frac{3+a_0}{48}\\8a_4-a_3=\frac16\implies a_4=\frac{1+6a_3}{48}=\frac{8+1+a_0}{384}=\frac{9+a_0}{384}$$... and we're done :-)

Answer 17

@oldrin.bataku shouldn't the first one just be
a0/2
?

Answer 18

or am I missing something?

Answer 19

oh damn I made a stupid error once again.. no big deal, we just end up with:$$a_1=\frac{a_0}2,a_2=\frac{2+a_0}8,a_3=\frac{2+a_0}{48},a_4=\frac{8+a_0}{384}$$

Answer 20

oh shoot \(a_4=\frac{8+2+a_0}{384}=\frac{10+a_0}{384}\)