What happens when a boundary condition is included?

[cytochrome]

Homework Statement

Show that sines and cosines are the solutions of the differential equation

f''(x) = (-σ^2)f(x)

What if a boundary condition is included that g(0) = 0?

Homework Equations

f''(x) = (-σ^2)f(x)

The Attempt at a Solution

Plugging in sin(σx) and cos(σx) yields an equality therefore the expression is true.

I'm just confused about the boundary condition.

If g(0) = 0 then only the sin(σx) works, correct?

 

[grzz]

I am assuming that g(0) = 0 stands for f(0) = 0.

The solution f(x) = sin(σx) satisfies the condition that f(0) = 0 for all values of σ. But the solution f(x) = cos(σx) satisfies the given condition only for particular values of σ.

 

[SammyS]

Homework Statement

Show that sines and cosines are the solutions of the differential equation

f''(x) = (-σ^2)f(x)

What if a boundary condition is included that g(0) = 0?

Homework Equations

f''(x) = (-σ^2)f(x)

The Attempt at a Solution

Plugging in sin(σx) and cos(σx) yields an equality therefore the expression is true.

I'm just confused about the boundary condition.

If g(0) = 0 then only the sin(σx) works, correct?

I hope you mean f(0) = 0 .

Yes, if f(0) = 0, then only sin(σx) works.

cos(σx) does not work for that boundary condition.

I am assuming that g(0) = 0 stands for f(0) = 0.

The solution f(x) = sin(σx) satisfies the condition that f(0) = 0 for all values of σ. But the solution f(x) = cos(σx) satisfies the given condition only for particular values of σ.

@grzz,

For what value of σ will cos(σ∙0) = 0 ?

 

[grzz]

Thanks SammyS for pointing out my mistake.

The last part of my post i.e.'But the solution f(x) = cos(σx) satisfies the given condition only for particular values of σ' is not correct.