# y=3e^{3x} is a solution of a second order linear homogeneous differential equation with constant coefficients. The equation is: (a) y''-(3+a)y'+3ay=0, a any real number. (b) y''+y'-6y=0 (c) y''+3y'=0 (d) y''+(3-a)y'+3ay=0, a any real number. (e) Cannot be determined.

Question
$$y=3e^{3x}$$ is a solution of a second order linear homogeneous differential equation with constant coefficients. The equation is:
(a) $$y''-(3+a)y'+3ay=0$$, a any real number.
(b) $$y''+y'-6y=0$$
(c) $$y''+3y'=0$$
(d) $$y''+(3-a)y'+3ay=0$$, a any real number.
(e) Cannot be determined.

2021-02-13
The given solution of a second order linear differential equation is $$y=3e^{3x}$$
Evaluate y' as follows.
$$y'=\frac{d}{dx}(3e^{3x})$$
$$=3(e^{3x}\cdot3)$$
$$=9e^{3x}$$
Hence, $$y'=9e^{3x}$$
Evaluate y'' as follows.
$$y''=\frac{d}{dx}(9e^{3x})$$
$$=9(e^{3x}\cdot3)$$
$$=27e^{3x}.$$
So, $$y''=27e^{3x}.$$
Now, substitute the values of y' and y'' in all the options and verify.
Option (a):
$$y''-(3+a)y'+3ay=0$$
$$(27e^{3x})-(3+a)(9e^{3x})+3a(3e^{3x})=0$$
$$27e^{3x}-27e^{3x}-9ae^{3x}+9ae^{3x}=0$$
$$0=0$$
Hence, option (a) satisfies the given solution.
Option (b):
$$y''+y'-6y=0$$
$$27e^{3x}+9e^{3x}-6(3e^{3x})=0$$
$$27^{3x}+9e^{3x}-18e^{3x}=0$$
$$18e^{3x}\neq0$$
Thus, option (b) does not satisfy the given solution.
Option (c):
$$y''+3y'=0$$
$$27e^{3x}+3(9e^{3x})=0$$
$$27e^{3x}+27e^{3x}=0$$
$$54e^{3x}\neq0$$
So, option (c) does not satisfy the given solution.
Option (d):
$$y''+(3-a)y'+3ay=0$$
$$(27e^{3x})+(3-a)(9e^{3x})+3a(3e^{3x})=0$$
$$27e^{3x}+27e^{3x}-9ae^{3x}+9ae^{3x}=0$$
$$64e^{3x}\neq0$$
Hence, option (c) does not satisfy the given solution.
Since, the only option that satisfied is (a), the correct option is (a).
Therefore, the answer is option (a).

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