# Thread: 2nd order diff equation!

1. ## 2nd order diff equation!

Hi, I have this differential equation to solve its broken down in to 3 parts but im having trouble doing the 2nd part any ideas would be great!

Consider the intial value problem

(y x^2) - (4xy) + 6y = g(x) x>0 y(0)=y(0)=0

part a) Write this de in self adjoint form which i have done

(1/x^4 y) + 6y/x^6 = g(x)x^-6

part b) Find the influence function for this

i know the formula for this but i have to find y1 and y2 but i have no idea on how to do this can anyone help me?

thanks

$\displaystyle \displaystyle {y=x^r }$

3. Originally Posted by FunkyBudduh
(y x^2) - (4xy) + 6y = g(x) x>0 y(0)=y(0)=0
put $\displaystyle t=\ln x$ then $\displaystyle \displaystyle\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{dt}{dx}=\frac{1}{x}\frac{dy}{dt},$ then $\displaystyle \displaystyle\frac{{{d}^{2}}y}{d{{t}^{2}}}-5\frac{dy}{dt}+6y(t)=g({{e}^{t}}).$

homogeneous solution is $\displaystyle y_h(t)=c_1e^{3t}+c_2e^{2t}$ now find the particular solution.

4. Hi Krizalid,

Why is it -5dy/dx?

thanks

5. it's dt not dx.

6. Hi Krizalid,

Why are we using the function t=ln(x) instead of y=x^r seeming as the LHS of the equation is a polynomial and not a exponential. When i have used y= x^r i have got y=x^2 and y=x^3.

thanks

7. it was a matter of choice, both ways work.

8. Hi Krizalid,

part 3 of the question is that $\displaystyle g(x) = (x^4)(e^x)$

i have used $\displaystyle y=x^2$ and $\displaystyle y=x^3$
to obtain a really nasty general solution, with exponentials and powers of xs

it looks like

$\displaystyle e^x(-x^7 + 8x^6 - 48x^5.....$

pretty sure this is wrong though, any idea on how to solve this i been tryin to do this for days now ...

thanks

much appreciated

9. I HAVE SOLVED THIS!!!!!!!!!!!!!!! Thx For your help all