Question #297976

Construct a h********** linear differential equation with constant coefficients whose solution is y = 6+3xe^x–cosx


1
Expert's answer
2022-02-15T17:25:20-0500

Let us construct a homogeneous linear differential equation with constant coefficients whose solution is y=6+3xexcosx.y = 6+3xe^x–\cos x.

It follows that the characteristic equation of this differential equation should be the following:

k(k1)2(k2+1)=0.k(k-1)^2(k^2+1)=0. Indeed, in this case the general solution is y=C1+(C2+C3x)ex+C4cosx+C5sinx,y=C_1+(C_2+C_3x)e^x+C_4\cos x+C_5\sin x, and hence we get the solution y=6+3xexcosxy = 6+3xe^x–\cos x when C1=6, C2=0, C3=3, C4=1, C5=0.C_1=6,\ C_2=0,\ C_3=3,\ C_4=-1,\ C_5=0.

The characteristic equation is equvalent to

(k32k2+k)(k2+1)=0,(k^3-2k^2+k)(k^2+1)=0,

and hence to

k52k4+2k32k2+k=0.k^5-2k^4+2k^3-2k^2+k=0.

Therefore, a homogeneous linear differential equation with constant coefficients whose solution is y=6+3xexcosxy = 6+3xe^x–\cos x is the following:

y(V)2y(IV)+2y(III)2y+y=0.y^{(V)}-2y^{(IV)}+2y^{(III)}-2y''+y'=0.


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