Answer 1

Given that, 

`x^2( d^2y)/dx^2+3x dy/dx+3y=(log x.cos (log x))/x` 

Putting z = log x and x = e^z, we get 

`[D(D-1)+3D+3]y=e^-z.z. cos z` 

`[D^2+2D+3]y=e^-z.z.cos z` 

∴ ` "The A.E is" D^2+2D+3=0` 

∴ `D= (-2+-2sqrt2.i)/2=-1+- sqrt2.i` 

∴`" The C.F. is y"= e^-z.z. cos z` 

=`e^-z 1/((D-1)^2+2(D-1)+3).z.cos z= e^-z. 1/(D^2+2).z.cos z`  

= `e^-z [Z-1/(D^2+2).2D].1/(D^2+2). cos z` 

=`e_z [z-1/(D^2+2).2D]cos z=e^-z [z cosz+1/(D^2+2).2 sin z]` 

=` e^-z[z cos z+2 sin z]` 

The complete solution is, 

`y=C.F.+P.I.` 

`y=e^-z (C_1 cos sqrt2z+C_2 sin sqrt2z)+e^-z[z cos z+2 sin z]` 

`Y=1/x(C_1 cos sqrt2 log x+C_2 sin sqrt2 log)+1/x [log x cos log x+2 sin log x]`