Question

x^2 dy/dx +2xy=1

Answer 1

\[y=\sum_0a_nx^n\]
\[y'=\sum_1a_nnx^{n-1}\]
sub and solve

Answer 2

this actually a differential equation

Answer 3

\[x^2 \sum_1a_nnx^{n-1} +2x\sum_0a_nx^n=1\]
\[ \sum_1a_nnx^{n+1} +\sum_02a_nx^{n+1}=1\]
\[ \sum_1a_n(n+2)x^{n+1} +2a_0~x=1\]
\[ \sum_1a_n(n+2)x^{n+1} =1-2a_0~x\]
maybe

Answer 4

the =1 at the end has me a little perplexed :) never tried this method for a nonhomogenous setup

Answer 5

can you explain?

Answer 6

i have done a method that is prolly a bit advanced from your first course diffyQ stuff

Answer 7

i defined a general power series as a solution and then attempt to solve the power series

Answer 8

could you come down to my level of first course diffQ stuff?

Answer 9

the solution itself can be determined by dividing off the x^2, adn subbing in an appropriate e^(rx) setup

Answer 10

and then?

Answer 11

\[x^2 \frac{ dy }{ dx }+2xy =1\]Is of the form:
\[y' +P(x)y=Q(x)\]Which means it is a first order linear differential equation.

First step is to get your y' term isolated, by dividing by x^2:
\[\frac{ dy }{ dx }+\frac{ 2 }{ x }y=\frac{ 1 }{ x^2 }\]
Next step is to construct an integrating factor:
\[\mu(x)=e ^{\int\limits_{}^{}P(x)dx}=e ^{\int\limits_{}^{}\frac{2}{x}dx}=e ^{2ln(x)}=e ^{ln(x^2)}=x^2\]
Then multiply the integrating factor through all your terms of your differential equation:
\[\ d[(x^2)(y(x))]=\int\limits_{}^{}\frac{ 1 }{ x^2 }x^2dx\]\[y(x)=\frac{\int\limits_{}^{}dx }{ x^2 }=\frac{ x }{ x^2 }+C=\frac{ 1 }{ x }+C\]
Anybody want to check me on this? I've never done this before, I just followed a formula.

Answer 12

solving for the =0 part first, then backing into the =1 setup

x^2 y' +2xy = 0

y' +2/x y = 0

get this into the form:

e^u y' + u' e^u y = 0

such that u' = 2/x
therefore u = 2ln(x)

Answer 13

Pompeii did an excellent write up of that mathod :)

Answer 14

Just as a mistake, I forgot the integral on the left-hand side of the equation in the second-to-last step. You need to integrate both sides to counter taking the derivative.

Answer 15

yh thanx guys i'll try to complete it. i like pompei's method.