Respuesta :
Answer:
k = 2
Step-by-step explanation:
As given,
the differential equation is - y'' + 9y = 26[tex]e^{-2x}[/tex] .......(1)
Let the solution of the differential equation be y = [tex]y_{c} + y_{p}[/tex]
The auxiliary equation becomes
m² + 9 = 0
⇒m² = -9
⇒m = ±√-9
⇒m = ± 3i
So, the complimentary solution becomes
[tex]y_{c}(x)[/tex] = A cos(3x) + B sin(3x)
Now,
Let the particular solution be [tex]y_{p}[/tex] = A[tex]e^{-2x}[/tex]
⇒[tex]y'_{p}[/tex] = -2A[tex]e^{-2x}[/tex]
and [tex]y''_{p}[/tex] = 4A[tex]e^{-2x}[/tex]
Now,
equation (1) becomes
4A[tex]e^{-2x}[/tex] + 9 ( A[tex]e^{-2x}[/tex] )= 26[tex]e^{-2x}[/tex]
⇒4A[tex]e^{-2x}[/tex] + 9A[tex]e^{-2x}[/tex] = 26[tex]e^{-2x}[/tex]
⇒13A[tex]e^{-2x}[/tex] = 26[tex]e^{-2x}[/tex]
By comparing , we get
13A = 26
⇒A = [tex]\frac{26}{13}[/tex] = 2
∴ we get
[tex]y_{p}[/tex] = 2[tex]e^{-2x}[/tex]
so, the solution becomes
y = A cos(3x) + B sin(3x) + 2[tex]e^{-2x}[/tex]
As given , the solution y=k[tex]e^{-2x}[/tex] +4cos(3x)
So , by comparing with the solution , we get k = 2
So, the correct option is (A) 2.
