Question

8. For what value of k, if any, will y=ke?' +4cos(3x)be a solution to the differential equation

y"+9y = 26e 22

(A) 2

(B)

13

5

(C) 26

(D) There is no such value of k.

Answer 1

k = 2

Explanation:

As given,

the differential equation is - y'' + 9y = 26
`e^(-2x)` .......(1)

Let the solution of the differential equation be y =
`y_(c) + y_(p)`

The auxiliary equation becomes

m² + 9 = 0

⇒m² = -9

⇒m = ±√-9

⇒m = ± 3i

So, the complimentary solution becomes

`y_(c)(x)` = A cos(3x) + B sin(3x)

Now,

Let the particular solution be
`y_(p)` = A
`e^(-2x)`

⇒
`y'_(p)` = -2A
`e^(-2x)`

and
`y''_(p)` = 4A
`e^(-2x)`

Now,

equation (1) becomes

4A
`e^(-2x)` + 9 ( A
`e^(-2x)` )= 26
`e^(-2x)`

⇒4A
`e^(-2x)` + 9A
`e^(-2x)` = 26
`e^(-2x)`

⇒13A
`e^(-2x)` = 26
`e^(-2x)`

By comparing , we get

13A = 26

⇒A =
`(26)/(13)` = 2

∴ we get

`y_(p)` = 2
`e^(-2x)`

so, the solution becomes

y = A cos(3x) + B sin(3x) + 2
`e^(-2x)`

As given , the solution y=k
`e^(-2x)` +4cos(3x)

So , by comparing with the solution , we get k = 2

So, the correct option is (A) 2.