Question

equation:x^(2)y'-2xy=cos(1/x)
I have to solve it using legrand's variation method or integral factor method.Could someone help me,please?

Answer 1

First arrange it in standard form so y' stands alone /:x^2
y'-2y/x=cos(1/x) / x^2

Answer 2

Than the integrating factor is
exp(integral -2/x)=exp(-2lnx)=x^-2

Answer 3

Multiplying with it will give you
y'x^-2-2yx^-3=cos(1/x)x^-4

Answer 4

Here you notice that
y'x^-2-2yx^-3=(yx^-2)'

Answer 5

So (yx^-2)'=cos(1/x)x^-4 pr
yx^-2=integral cos(1/x)x^-4

Answer 6

Now I have to take out my pizza from the oven .-)

Answer 7

now Im stuck a bit. We need to find the integral cos(1/x)x^-4

Answer 8

Sorry I have no idea at all how to do that.

Answer 9

ok, i m starting to look your soultion

Answer 10

Here comes the cavalry! :-)

Answer 11

:) but you re saying that x^(-2) is the integral factor ?

Answer 12

Yes. I dont know legrand's method. Maybe amistre is doing that right now

Answer 13

hmm,you re right it is the integral factor,and we have to integrate cos(1/x)/x^4,which is imposible to inte
grate with elementary functions :(

Answer 14

\[x^2y'-2xy=cos(1/x)\]
\[\frac{1}{x^2}(x^2y'-2xy=cos(1/x))\]
\[y'-\frac{2}{x}y=\frac{cos(1/x)}{x^2}\]
\[exp(-2ln|x|)\ (y'-\frac{2}{x}y=\frac{cos(1/x)}{x^2})\]
\[exp(-2ln|x|) y'+\ \frac{2}{x}exp(-2ln|x|)\ y=exp(-2ln|x|)\frac{cos(1/x)}{x^2}\]
\[exp(-2ln|x|) y\ =\ \int exp(-2ln|x|)\frac{cos(1/x)}{x^2}\ dx\]
\[y\ =exp(2ln|x|) \int exp(-2ln|x|)\frac{cos(1/x)}{x^2}\ dx+C_1exp(-2ln|x|)\]
\[y\ =x^2 \int \frac{cos(1/x)}{x^4}\ dx+C_1x^{-2}\]

Answer 15

x^4 can be partial fractioned i believe

Answer 16

\[\frac{cos(1/x)}{x^4}=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{x^3}+\frac{d}{x^4}\]
\[cos(1/x)=ax^3+bx^2+cx+d\]
maybe
or some other trick

Answer 17

integrate by parts perhaps

Answer 18

ou,yes , i had forgotten of partial fractions

Answer 19

\[\int\frac{1}{x^2}\frac{cos(\frac{1}{x})}{x^2}dx\]
\[=-\frac{1}{x^2}sin(\frac{1}{x})-\int \frac{2}{x^3}sin(\frac{1}{x})dx\]
\[=-\frac{1}{x^2}sin(\frac{1}{x})-2\int \frac{1}{x}\frac{sin(\frac{1}{x})}{x^2}dx\]
\[=-\frac{1}{x^2}sin(\frac{1}{x})-2(-\frac{1}{x}cos(\frac{1}{x})+\int \frac{}{})\]
oy this gives me a head ache ...

its gonna be a mess of an integration by parts ...

Answer 20

http://www.wolframalpha.com/input/?i=integrate+cos%281%2Fx%29%2Fx^4

Answer 21

i dont know the other method either ...

Answer 22

ok,thanks,the other method in the end has the same integration problem.