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.

Second order linear equations
ANSWERED
asked 2021-02-12
\(y=3e^{3x}\) 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.

Answers (1)

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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