Answer to Question #297976 in Differential Equations for Santoshi

Question

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

Answer 1

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

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

$k(k-1)^2(k^2+1)=0.$ Indeed, in this case the general solution is $y=C_1+(C_2+C_3x)e^x+C_4\cos x+C_5\sin x,$ and hence we get the solution $y = 6+3xe^x–\cos x$ when $C_1=6,\ C_2=0,\ C_3=3,\ C_4=-1,\ C_5=0.$

The characteristic equation is equvalent to

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

and hence to

$k^5-2k^4+2k^3-2k^2+k=0.$

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

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