Question

Solve by method of variation of parameters :`(D^2-6D+9)y=e^(3x)/x^2`

Answer 1

`(D^2-6D+9)y=e^(3x)/x^2`
For complementary solution ,
𝒇(𝑫)=𝟎

`therefore(D^2-6D+9)=0`

Roots are : D = 3 , 3 Real roots but repeatative.
The complementary solution of given diff. eqn is ,

`therefore y_c=(c_1+xc_2)e^(3x)`

For particular solution ,
By method of variation of parameters,

`y_p=y_1p_1+y_2p_2`

where `p_1=int(-y_2X)/wdx`

`p_2=int(-y_1X)/wdx`

`w=|(y_1,y_2),(y'_1,y'_2)|`

`w=|(e^(3x),xe^(3x)),(3e^(3x),e^(3x)+3xe^(3x))|=e^(6x)`

`p_1=int(-y_2X)/wdx=int(xe^(3x))/e^(6x).e^(3x)/x^2dx=int(-1)/xdx=-logx`

`p_2=int(-y_1X)/wdx=int(e^(3x))/e^(6x).e^(3x)/x^2dx=int(1)/x^2dx=(-1)/x`

The particular integral of given diff. eqn is given by,

`thereforey_p=-e^(3x)logx-e^(3x)=-e^(3x)(logx+1)`

The general solution of given diff. eqn is given by ,

`y_g=y_c+y_p=(c_1+xc_2)e^(3x)=-e^(3x)(logx+1)`