Question #220880
Solve (D2 + a2)y = sinax
1
Expert's answer
2021-07-27T17:12:04-0400

Let us solve the differential equation (D2+a2)y=sinax.(D^2 + a^2)y = \sin ax. The characteristic equation k2+a2=0k^2+a^2=0 has the solutions k1=ai,k2=ai,k_1=ai, k_2=-ai, and hence the general solution of the homogeneous differential equation is y=C1cosax+C2sinax.y=C_1\cos ax+ C_2\sin ax. The particular solution of non-homogeneous equation is of the form yp=x(Lsinax+Mcosax).y_p=x(L\sin ax+ M\cos ax). Then yp=Lsinax+Mcosax+x(aLcosaxaMsinax),y_p'=L\sin ax+ M\cos ax+x(aL\cos ax-a M\sin ax), and yp=aLcosaxaMsinax+aLcosaxaMsinax+x(a2Lsinaxa2Mcosax).y_p''=aL\cos ax-a M\sin ax+aL\cos ax-a M\sin ax+x(-a^2L\sin ax-a^2 M\cos ax).

It follows that 2aLcosax2aMsinax+x(a2Lsinaxa2Mcosax)+a2x(Lsinax+Mcosax)=sinax,2aL\cos ax-2a M\sin ax+x(-a^2L\sin ax-a^2 M\cos ax)+a^2x(L\sin ax+ M\cos ax)=\sin ax,

and hence 2aL=02aL=0 and 2aM=1.-2aM=1. Therefore, if a0a\ne 0 then L=0,M=12a.L=0,M=-\frac{1}{2a}.


We conclude that if if a0a\ne 0 then the general solution of the differential equation (D2+a2)y=sinax(D^2 + a^2)y = \sin ax is y=C1cosax+C2sinax12axcosax.y=C_1\cos ax+ C_2\sin ax-\frac{1}{2a}x\cos ax. If a=0a=0 then the general solution of the differential equation D2y=0D^2 y = 0 is y=C1x+C2.y=C_1x+C_2.


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