Question

find the general formula of
(dy/dx)+((2x+1)/x)*y=e^(-2x). y(1)=1/(e^2)

(i used ' ^ ' as power sign.)

Answer 1

\[\large\rm y'+\left(\frac{2x+1}{x}\right)~y=e^{-2x},\qquad\qquad y(1)=\frac{1}{e^2}\]

Answer 2

First order, linear,
so you'll want to find an integrating factor, ya? :)

Answer 3

i want to find y(x) but yes we will use integrating factor

Answer 4

I.F\[=e ^{ \int\limits \frac{ 2x+1 }{ x }dx}=?\]

Answer 5

i want to find the general solution of y(x)

Answer 6

I would probably recommend simplifying this fraction before integrating :)\[\large\rm \frac{2x+1}{x}\qquad=\frac{2x}{x}+\frac{1}{x}\qquad=2+\frac{1}{x}\]That's a little easier to integrate, ya?

Answer 7

i already did that

Answer 8

i also found a general solution but it does not satisfy y(1)=1/(e^2)

Answer 9

So what do you get for your integrating factor? :d

Answer 10

e^(2x) * x

Answer 11

Ah good good good.
Multiplying through by our integrating factor,\[\large\rm x e^{2x}y'+e^{2x}(2x+1)y=1\]I simplied already, hopefully you understood that,
otherwise I can back up a step.
(Don't forget to give the RIGHT SIDE your integrating factor as well)

Answer 12

Assuming our integrating factor is correct,
left side of the equation should reverse product rule,
and sure enough, it does here :)\[\large\rm \left(yxe^{2x}\right)'=x\]Woops I messed up the right side on the previous post, sorry about that.

Answer 13

\[C.S.~is ~y*xe^{2x}=\int\limits e ^{-2x}x e ^{2x}dx=\frac{ x^2 }{ 2 }+c\]
when x=1 \[y=\frac{ 1 }{ e^2 }\]
\[\frac{ 1 }{ e^2 }*1*e^2=\frac{ 1^2 }{ 2 }+c,c=1-\frac{ 1 }{ 2 }=\frac{ 1 }{ 2 }\]
substitute the value of c and find the general solution.

Answer 14

ok, thank you very much

Answer 15

yw