# ED

#### Ale_M

Can someone help me with this problem, but using the parameter variation method, plis?

x^2y"-2xy'+2y=6x^3e^2x

#### idontknow

let $$\displaystyle c_1 y_1 ,c_2 y_2$$ be solutions of $$\displaystyle x^2y_h'' -2xy'_h+2y_h =0$$ . (can you continue about $$\displaystyle y_h$$ ?)

$$\displaystyle y_p = - {y_1}\int{{\dfrac{{{y_2}6x^3e^2x }}{{W\left( {{y_1},{y_2}} \right)}}\,dx}} + {y_2}\int{{\dfrac{{{y_1}6x^3e^2x }}{{W\left( {{y_1},{y_2}} \right)}}\,dx}}$$ ; where $$\displaystyle \; \; W(y_1 , y_2 )=y_1 'y_2 -y_1 y_2 '$$.
the general solution is $$\displaystyle y=y_h +y_p$$.

topsquark

#### idontknow

set $$\displaystyle x=e^t$$ ; $$\displaystyle \; x^2y_h'' -2xy'_h+2y_h =e^{2t} \frac{d}{de^t} \left( \frac{d}{de^t} y_h \right)-2e^t \frac{dy_h }{de^t} +2y_h =e^t \frac{d}{dt}\left( e^{-t}\frac{dy_h }{dt}\right) -2\frac{dy_h}{dt}+2y_h =e^t [ e^{-t}\frac{d^2 y_h }{dt^2 }-e^{-t}\frac{dy_h }{dt}]-2\frac{dy_h }{dt} +2y_h =0.$$

$$\displaystyle \frac{d^2 y_h }{dt^2 }-3\frac{dy_h }{dt} +2y_h =0$$ ; $$\displaystyle \; y_h =e^{rx} \implies r^2 -3r+2=(r_1 -1)(r_2 -2)=0$$ ; $$\displaystyle \; (r_1 , r_2 ) = (1 , 2 )$$.
$$\displaystyle y_h = C_1 e^x +C_2 e^{2x}$$ ; $$\displaystyle \; (y_1 , y_2 ) = (C_1 e^x , C_2 e^{2x} )$$.

topsquark

#### skipjack

Forum Staff
x^2y"-2xy'+2y=6x^3e^2x
I think you intended $x^2y'' - 2xy' + 2y = 6x^3e^{2x}$.

If you did, the solution will be $y = \text{C}_1x^2 + \text{C}_2x + \large\frac32xe^{2x}\!$.

Unfortunately, idontknow got careless, to say the least.

By the way, the solution is very easy to find by using $x^{-3}$ as an integrating factor.

#### idontknow

Edit
set $$\displaystyle x=e^t$$ ; $$\displaystyle \; x^2y_h'' -2xy'_h+2y_h =e^{2t} \frac{d}{de^t} \left( \frac{d}{de^t} y_h \right)-2e^t \frac{dy_h }{de^t} +2y_h =e^t \frac{d}{dt}\left( e^{-t}\frac{dy_h }{dt}\right) -2\frac{dy_h}{dt}+2y_h =e^t [ e^{-t}\frac{d^2 y_h }{dt^2 }-e^{-t}\frac{dy_h }{dt}]-2\frac{dy_h }{dt} +2y_h =0.$$

$$\displaystyle \frac{d^2 y_h }{dt^2 }-3\frac{dy_h }{dt} +2y_h =0$$ $$\displaystyle \implies r^2 -3r+2=(r_1 -1)(r_2 -2)=0$$ ; $$\displaystyle \; (r_1 , r_2 ) = (1 , 2 )$$.
$$\displaystyle y_h = C_1 e^t +C_2 e^{2t}$$ ; $$\displaystyle \; (y_1 , y_2 ) = (C_1 x , C_2 x^2 )$$.

Last edited: