Question

Solve differential equation y''-k^2*y=0 with initial conditions y'(0)=-y and y'(-d)=0. Thanks

Answer 1

can you use leplace yet?

Answer 2

or integration method?
let u = e^int(-k^2)dt

Answer 3

\[y''-k^2y=0\qquad\Longrightarrow\qquad y(x)=A\sinh(kx)+B\cosh(kx)\]

Answer 4

@aliyuaziz you mean y'(0)=-y(0)?

Answer 5

my issue is with the boundary condition which contains y:
y'(0) = -y.

Answer 6

can't use laplace yet please

Answer 7

i haven't seen boundary condition like that!!!

Answer 8

This is my first time to encounter that type of boundary condition too.

Answer 9

can you find \[y^\prime(x)\]

Answer 10

y'(x) = Akcosh(kx) + Bksinh(kx)

Answer 11

now you can substitute those boundary conditions

Answer 12

i think that is not a correct type of boundary condition!!!

Answer 13

According to UnkleRhaukus, y(x) = Asinh(kx) + Bcosh(kx)
y'(x) = Akcosh(kx) + Bksinh(kx)
at y'(0) = -y, A = -y/k and at y'(-d) = 0, B = y/(k*tanh(-kd)
therefore y(x) = -y/k*sinh(kx) + y/k*cosh(kx)/tanh(-kd)

correct?

Answer 14

you treated y like a constant that is not correct

Answer 15

in Ordinary Differential Equations we have not boundary condition like that! could u plz check the question again? maybe it is y'(0)=-y(0)

Answer 16

or something else

Answer 17

@mukushla, thanks for pointing out the treatment of y as a constant which is wrong. The issue is on how do we solve solve problems where the boundary condition is not constant but dependent on the solution?

Answer 18

let me say something

Answer 19

if y'(0)=-y then y=-(A+B) so y is constant then y'=0 and y''=0 and the equation will be -k^2y=0 so y=0!!!!

Answer 20

err y=-B