Question

Which is not a solution to the differential equation y"+4y'=0?

A. y = 10
B. y = 4e⁻²ˣ
C. y = 3sin (2x)
D. y = 2 cos(2x)+ 4

Answer 1

Option C, y = 3sin(2x), and option D, y = 2cos(2x) + 4, do not satisfy the differential equation y" + 4y' = 0.

Step-by-step explanation:

To determine which option is not a solution to the differential equation y" + 4y' = 0, we can test each function by taking its derivatives and substituting them into the equation.

For A. y = 10,

the first derivative y' = 0 and

the second derivative y" = 0.

Substituting these into the differential equation gives 0 + 4(0) = 0 which satisfies the equation.

For B. y = 4e⁻²ˣ,

y' = -8e⁻²ˣ and

y" = 16e⁻²ˣ.

Substituting into the differential equation gives 16e⁻²ˣ + 4(-8e⁻²ˣ) = 0, which also satisfies the equation.

For C. y = 3sin(2x),

y' = 6cos(2x) and

y" = -12sin(2x).

Substituting into the differential equation gives -12sin(2x) + 4(6cos(2x)) = 24cos(2x) - 12sin(2x) which does not satisfy the equation as it does not simplify to zero.

Finally for D. y = 2cos(2x) + 4,

y' = -4sin(2x) and

y" = -8cos(2x).

Substituting into the differential equation gives -8cos(2x) + 4(-4sin(2x)) = -8cos(2x) - 16sin(2x) which does not satisfy the equation as it does not simplify to zero.

Therefore, the functions that do not satisfy the differential equation are C. y = 3sin(2x) and D. y = 2cos(2x) + 4.