Recent content by Thumper88

I can do that...lol. Thanks a lot.

that is correct, right? just checking, thanks for all the help!

been a while since I've done any integration by parts... ok...so with u(x) = - \frac {e^{-x}}{x} y_2(x) = - \frac {e^{-x}}{x} * e^x = -\frac {1}{x} Sooo... y(x) = C_1 e^x + C_2 (-x^{-1})

the -4e^-x came from using a minus sign instead of the negative key...

arg...made some typos on my calculator...not getting those answers anymore.

u(x) = \int w dx u(x) = \int \frac {(x+1) e^-x}{x^2} dx = \int \frac {e^{-x}}{x} dx + \int \frac {e^{-x}}{x^2} dx \int u dv = uv - \int v du \int \frac {e^{-x}}{x} dx u = \frac {1}{x} du = -\frac {1}{x^2}dx dv = e^{-x} dx v = -e^{-x} \int u v' = \frac...

so would the general solution just be: y(x) = C_1 e^x + C_2 Where -4 is included in the constant C_2 .

u(x) = -4 e^{-x} so y_2(x) = -4 e^{-x} * e^x = -4 correct?

yea...i was editing the equation...still getting used to the code

I redid the algebra by hand and got exactly what you got. Separating the variables -\frac {1}{w} dw = \frac {(x^2+2x+2)}{(x^2+x)}dx integrating both sides and solving for w I get: w = \frac {((x+1) * e^-x)}{x^2} dx So now u = integral of w dx

yea, I'm at a halt here..really not sure what to do next.

ok, well...to further things a little... I distributed everything out and added like terms, leading to: 3u'x^2e^x + 2u'xe^x - 2u'e^x + u''x^2e^x + u''xe^x = 0 Using the variable transformation, w = u' and w' = u'' 3wx^2e^x + 2wxe^x - 2we^x + w'x^2e^x + w'xe^x = 0 Still...

So I have the derivatives (u = v(x)): y_2 = u y_1 = ue^x y_2' = uy_1' + y_1u' = ue^x + e^xu' y_2'' = uy_1'' + 2u'y_1' + y_1u'' = ue^x + 2u'e^x + e^xu'' Putting these into the original DE... (x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0 (x^2 +...

Homework Statement Consider the differential equation: (x^2 + x) \frac {d^2y} {dx^2} - (x^2 - 2) \frac {dy} {dx} - (x + 2)y = 0 Given that y = e^x is a solution, what is the general solution of this equation? Homework Equations ?The Attempt at a Solution I don't know where to...