Question

Diffs Question @mathematics

Answer 1

solve: \[x^2 y'' + y' = 4x \]

Answer 2

try using bernoulli

Answer 3

okay I have tried using the substitution p= y' = dy/dx ; dp/dx = y'' but ill take a look at that

Answer 4

i dont think bernoulis works because we would need a y^n

Answer 5

there may be a way to get it there but im not sure how

Answer 6

burnnoodle is for a "y" part on the right i thought, or am i confusing names again

Answer 7

=D

Answer 8

divide of the x^2; and consider y'' and y' to be renamed as z' and z

Answer 9

okay did that

Answer 10

your just solving for a first order by looks then

x^2 y'' + y' = 4x

x^2 z' + z = 4x

z' + x^-2 z = 4x^-1

{S} x^-2 dx = -x^-1

e^-1/x z = {S} 4x^-1 ....yada yada

Answer 11

you solve for z then becasue z = y'

int (z=y') ; int(z) = y

Answer 12

i end up with this integral on the right hand side:

http://www.wolframalpha.com/input/?i=integrate+%28x*e^x^-1%29^-1

Answer 13

yeah, also see if itll go separable by chance

Answer 14

I tried but couldn't figure out a way to separate it

Answer 15

\[x^2\frac{dz}{dt}+z=4x \]
\[\frac{dz}{dt}+x^{-2}z=4x^{-1} \]
yeah, doesnt look like itll pull out nicely

Answer 16

I emailed my professor and this was our conversation:

Hi Eric,

Left with no other choice, we can express an anti-derivative for a function f(x) as the integral from a to x of f(t) dt + C. (You can choose any value for a, as long as you are careful about avoiding discontinuities of f.)

Does that help?

--Ben


On Sun, Feb 26, 2012 at 1:06 PM, Eric wrote:
Dr. Levitt

I'm having trouble with the first reducible differential equation on the homework. Im probably just making a mistake but will you take a quick look and make sure there was not a typo. When i use the integrating factor I end up with an integral that is difficult to solve.

The equation is:

x^2 y'' + y' = 4x

Thanks,

Eric

Answer 17

\[e^{-x^{-1}}z=\int 4x^{-1}e^{-x^{-1}}dx\]
\[e^{x}z=\int 4x^{-1}e^{x}dx\]
\[e^{x}z=4e^xln(x)-\int 4x^{-1}e^{x}dx\]

that might mean we can leave it as an integration for exactness as long as our interval is consistent

Answer 18

\[x^2 m' + m = 4x\]

\[x^2 m' + m = 0\]

\[m' +x^{-2} m = 0\]

\[e^{-x^{-1}} m = c_1\]

\[m = c_1e^{x^{-1}}\]
might be a homogenous solution

Answer 19

\[\int (e^{1/x}=y')\to y=\]might be a little tricky as well

Answer 20

we could turn it into a series expansion:

http://www.wolframalpha.com/input/?i=e%5E%281%2Fx%29

\[\int 1+\sum_{n=1}^{inf}\frac{1}{n!}x^{-n}dx\]
\[\sum_{n=0}^{inf}\frac{1}{n!(-n+1)}x^{-n+1}+C \]

just a thought, prolly a useless stray thought

Answer 21

might have to shave off one more term to get that off a x^-1 tho