Respuesta :
Answer:
c. 3
Step-by-step explanation:
y = k sin(5x) + 2 cos(4x)
y' = 5k cos(5x) − 8 sin(4x)
y" = -25k sin(5x) − 32 cos(4x)
y'' + 16y = -27 sin(5x)
(-25k sin(5x) − 32 cos(4x)) + 16(k sin(5x) + 2 cos(4x)) = -27 sin(5x)
-25k sin(5x) − 32 cos(4x) + 16k sin(5x) + 32 cos(4x) = -27 sin(5x)
-9k sin(5x) = -27 sin(5x)
k = 3
The value of k for the differential equation is 3.
Given that,
The equation y = k sin (5x) + 2cos (4x) be a solution to the differential equation y'' +16y =-27sin(5x).
We have to determine,
The value of k.
According to the question,
The equation y = k sin (5x) + 2cos (4x) be a solution to the differential equation y'' +16y =-27sin(5x).
Equation; y = k sin (5x) + 2cos (4x)
An equation of the form which is linear in y and its derivatives is called a second-order linear differential equation.
This method is used when the given differential equation can be written in the form of dy/dx = f(y)g(x), where the function f is the function of y only and the function g is the function of x only.
On differentiating the equation with respect to x,
[tex]\rm \dfrac{dy}{dx} = \dfrac{d(k sin (5x))}{dx} + \dfrac{d(2cos (4x) )}{dx}\\\\ \dfrac{dy}{dx} = k\dfrac{d(sin (5x))}{dx} + 2\dfrac{d(cos (4x) )}{dx}\\\\y' = 5k cos(5x) -8 sin(4x)[/tex]
Again differentiating with respect to x,
[tex]\rm \dfrac{dy'}{dx} = 5k \dfrac{(cos(5x)}{dx} -8\dfrac{(sin(4x))}{dx}\\\\y" = -25k sin(5x) - 32 cos(4x)[/tex]
On substitute the values in the given equation,
[tex]\rm y'' +16y =-27sin(5x)[/tex]
[tex]\rm -25k sin(5x) - 32 cos(4x) + 16(ksin (5x) + 2cos (4x)) = -27sin(5x)\\\\ -25k sin(5x) - 32 cos(4x) + 16ksin (5x) + 32cos (4x)= -27sin(5x)\\\\ - 32 cos(4x) + 16ksin (5x) + 32cos (4x)= -27sin(5x) \\\\-25ksin(5x)+16ksin (5x) =- 27sin(5x)\\\\-9k = -27\\\\k = \dfrac{-27}{-9}\\\\k = 3[/tex]
Hence, The value of k is 3.
To know more about Differentiation click the link given below.
https://brainly.com/question/18062027
