Question #c6898

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Answer 1 · 2017-12-10T18:29:55

$y=c_1*sin2x+c_2*cos2x+1/4xsin2x-1/12cos4x$

$(D^2+4)y=cos2x+cos4x$

Characteristic equation of differential equation $r^2+4=0$

Roots of it are $r_1=-2i$ and $r_2=2i$

Hence homogeneous part of it $y_h=c_1*sin2x+c_2*cos2x$

Due to $cos2x$ is root of homogeneous part of it, particular solution of it is $y_p=Axsin2x+Bxcos2x+Csin4x+Dcos4x$

Consequently,

$D^2(Axsin2x+Bxcos2x+Csin4x+Dcos4x)+4(Axsin2x+Bxcos2x+Csin4x+Dcos4x)=cos2x+cos4x$

$4Acos2x-4Axsin2x-4Bsin2x-4Bxcos2x-16Csin4x-16Dcos4x+4Axsin2x+4Bxcos2x+4Csin4x+4Dcos4x=cos2x+cos4x$

$4Acos2x-4Bsin2x-12Csin4x-12Dcos4x=cos2x+cos4x$

After equating coefficients,

$4A=1$ , $-4B=0$ , $-12C=0$ and $-12D=1$

So, $B=C=0$ , $A=1/4$ and $D=-1/12$

Consequently, $y_p=1/4xsin2x-1/12cos4x$

Thus, $y=y_h+y_p=c_1*sin2x+c_2*cos2x+1/4xsin2x-1/12cos4x$