# Thread: another DE problem (well more of a parts prob)

1. ## another DE problem (well more of a parts prob)

Me agian, catch up on my DE HW...(150ish DE's in 4days ) im doing a linear prob.

Question: x^2(dy/dx) + x(x+2)y = e^x

I have found roe (p) and now im trying to intergrate the right side....i come up with... Dx[2x(e^x)y] = 2 (intergal sign)(x^-1)(e^2x)

ive tryed to use intergrating by parts....but it helps not as i get down to (above intregal) (intregal sign) (-x^(-2))x((1/2x)e^2x) and im back where i started..

The answer i need to get it to is....y= 1/2(x^-2)e^x+cx^-2(e^-x)

I'd appreciate any help...also any hints to what i did wrong

2. $x^2 y' + x(x + 2)y = e^x \implies y' + x^{ - 1} (x + 2)y = x^{ - 2} e^x$

Now find the integrating factor.

3. i got to there....my intergrating factor came out to be....2x(e^x)

4. I think the integrating factor should be $e^xx^2$

Show your process to find it.

5. the biggest problem i get is when i go to intergrate the right side and use intergration by parts...i get u = x^-1 du= -x^-2 dv= e^2x V = 1/2 e^2x

but that brings me back to x^-1(1/2 e^2x) (intragal sign) (1/2e^2x)(x^-1)

which just gives me another parts intregal....almost the same one minus the constant multiplyer which wont really help anything

not really sure if i just doing my math wrong or which

6. (x+2)/x -> 1+2/x

e^(intregal) 1+2/x -> x+ 2ln lxl -> e and ln cancel leaving 2xe^x

7. $\exp \int {\frac{{x + 2}}
{x}~dx} = e^{x + 2\ln \left| x \right|} = e^x \cdot e^{\ln x^2 } = e^x x^2$

8. oh duh.....sorry, i havent used log, ln and "e" in ages....forgot all the rules. know a good place to brezze back up on em?

9. It's quite dangerous solving differential equations if you don't consider the basic properties.

I've got to go now, but I hope someone could give you a 'refresh'

Greetings