1. ## General solution step

I'm attepmting to solve a second order partial differential equation using the seperation of variable method, but have gotten up to where i have reduced to ordinary differential equations and has been so long since i did them i'm confused.

If i have say;

X'' - X/(k^2) = 0

And i try a solution of the form X = $e^{(\lambda x)}$

hence X' = $\lambda e^{(\lambda x)}$

and X'' = $\lambda^2 e^{(\lambda x)}$

when substituted in gives;

$\lambda^2 e^{(\lambda x)} - \frac{e^{(\lambda x)}}{k^2} = 0$

hence; $\lambda = + (1/k) or - 1/k$

is this correct up to here? and the following is where i'm confused,

this seems to me to be real distinct solutions hence general solution should be;

X(x) = $A.e^{\frac{x}{k}} +B.e^{\frac{-x}{k}}$

but i remember the lecturer saying that this should be a trig or hyperbolic solution, as there is no complex roots i'm gussing hyperbolics?

Any help would be greatly appreciated, i'm very hazy on this stuff,

Cheers.

2. Originally Posted by monster
I'm attepmting to solve a second order partial differential equation using the seperation of variable method, but have gotten up to where i have reduced to ordinary differential equations and has been so long since i did them i'm confused.

If i have say;

X'' - X/(k^2) = 0

And i try a solution of the form X = $e^{(\lambda x)}$

hence X' = $\lambda e^{(\lambda x)}$

and X'' = $\lambda^2 e^{(\lambda x)}$

when substituted in gives;

$\lambda^2 e^{(\lambda x)} - \frac{e^{(\lambda x)}}{k^2} = 0$

hence; $\lambda = + (1/k) or - 1/k$

is this correct up to here? and the following is where i'm confused,

this seems to me to be real distinct solutions hence general solution should be;

X(x) = $A.e^{\frac{x}{k}} +B.e^{\frac{-x}{k}}$

but i remember the lecturer saying that this should be a trig or hyperbolic solution, as there is no complex roots i'm gussing hyperbolics?

Any help would be greatly appreciated, i'm very hazy on this stuff,

Cheers.
I would agree with your solution if I knew that $k$ was real. But it might be complex.

If it was complex then you could write

$k = \alpha + i\beta$.

Therefore $\frac{x}{k} = \frac{x}{\alpha + i\beta} = \frac{x(\alpha - i\beta)}{\alpha^2 + \beta^2} = \frac{\alpha x - i\beta x}{\alpha^2 + \beta^2}$.

Now you can use Euler's Formula to clean it up.

3. yes k is a real number such that k > 0,

in my notes there is an example where;

X'' + (k^2) X = 0

When using trial solution $e^{\lambda x}$

gets to $\lambda = (+-) ki$

so general solution is;

X = $a. e^{kxi} + b. e^{-kxi}$

which goes to;

X = Acos(kx) + Bsin(kx)
which i don't understand as;

1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?

4. Originally Posted by monster
yes k is a real number such that k > 0,

in my notes there is an example where;

X'' + (k^2) X = 0

When using trial solution $e^{\lambda x}$

gets to $\lambda = (+-) ki$

so general solution is;

X = $a. e^{kxi} + b. e^{-kxi}$

which goes to;

X = Acos(kx) + Bsin(kx)
which i don't understand as;

1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?
Because $e^{i\theta} = \cos{\theta} + i\sin{\theta}$ - this is Euler's Formula.

5. Originally Posted by monster
yes k is a real number such that k > 0,

in my notes there is an example where;

X'' + (k^2) X = 0

When using trial solution $e^{\lambda x}$

gets to $\lambda = (+-) ki$

so general solution is;

X = $a. e^{kxi} + b. e^{-kxi}$

which goes to;

X = Acos(kx) + Bsin(kx)
which i don't understand as;

1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?
You have both $e^{kxi}$ and $=ke^{-kxi}$.

Yes, $cos(x)= \frac{e^{ix}+ e^{-ix}}{2}= cos(x)$ and you should also know that $sin(x)= \frac{e^{ix}- e^{-ix}}{2i}$.

Multiplying the second equation by i, $i sin(x)= \frac{e^{ix}- e^{-ix}}{2}$ and adding that to the first, $cos(x)+ i sin(x)= e^{ix}$, the formula Prove It gave. Subtracting that from the first gives $cos(x)- i sin(x)= e^{-ix}$.

From that, $ae^{ix}+ be^{-ix}= a(cos(x)+ i sin(x))+ b(cos(x)- i sin(x))$ $= acos(x)+ ai sin(x)+ bcos(x)- bi sin(x)$ $= (a+ b)cos(x)+ i(a- b)sin(x)= A cos(x)+ B sin(x)$ with A= a+ b and B= a- b. If a and b were equal, that would be only a "cos(x)" term. It is because they are not equal that you have both cos(x) and sin(x).