Question

 Evaluate (𝑫𝟐 + 𝟔𝑫 + 𝟗)𝒚 = 𝒆 −𝟑𝒕 − 𝟑𝒄𝒐𝒔𝟒𝒕 + 𝒍𝒐𝒈5

Answer 1

 Let us solve the differential equation $(D^2 + 6D + 9)y = e^{−3t} − 3\cos 4t +\log 5.$

The characteristic equation $k^2+6k+9=0$ is equivalent to $(k+3)^2=0,$ and hence has the roots $k_1=k_2=-3.$ It follows that the general solution is of the form $y=(C_1+C_2t)e^{-3t}+y_p,$

where $y_p=at^2e^{-3t}+b\cos 4t+c\sin 4t+d.$ Then $y'_p=2ate^{-3t}-3at^2e^{-3t}-4b\sin 4t+4c\cos 4t, y''_p=2ae^{-3t}-6ate^{-3t}-6ate^{-3t}+9at^2e^{-3t}-16b\cos 4t-16c\sin 4t.$

We get the equation $2ae^{-3t}-6ate^{-3t}-6ate^{-3t}+9at^2e^{-3t}-16b\cos 4t-16c\sin 4t+6(2ate^{-3t}-3at^2e^{-3t}-4b\sin 4t+4c\cos 4t)+9(at^2e^{-3t}+b\cos 4t+c\sin 4t+d)=e^{-3t}-3\cos 4t+\log 5,$

which is equivalent to

$2ae^{-3t}+(-7b-24c)\cos 4t+(-7c-24b)\sin 4t+9d=e^{-3t}-3\cos 4t+\log 5.$

We conlude that $2a=1,\ -7b-24c=-3,\ -7c-24b=0,\ 9d=\log 5.$ It follows that $a=\frac{1}2,\ d=\frac{1}9\log 5, c=-\frac{24}7b,\ 7b-\frac{576}7b=3.$ Consequently, $a=\frac{1}2,\ d=\frac{1}9\log 5, \ b=-\frac{21}{527}, c=\frac{72}{527}.$

We conclude that the general solution of the differential equation $(D^2 + 6D + 9)y = e^{−3t} − 3\cos 4t +\log 5$ is

$y=(C_1+C_2t)e^{-3t}+\frac{1}2t^2e^{-3t}-\frac{21}{527}\cos 4t+\frac{72}{527}\sin 4t+\frac{1}9\log 5.$