# General solution step

#### 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 =$$\displaystyle e^{(\lambda x)}$$

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

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

when substituted in gives;

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

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

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

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.

#### Prove It

MHF Helper
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 =$$\displaystyle e^{(\lambda x)}$$

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

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

when substituted in gives;

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

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

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

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 $$\displaystyle k$$ was real. But it might be complex.

If it was complex then you could write

$$\displaystyle k = \alpha + i\beta$$.

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

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

#### 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 $$\displaystyle e^{\lambda x}$$

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

so general solution is;

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

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 ?

#### Prove It

MHF Helper
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 $$\displaystyle e^{\lambda x}$$

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

so general solution is;

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

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 $$\displaystyle e^{i\theta} = \cos{\theta} + i\sin{\theta}$$ - this is Euler's Formula.

• monster

#### HallsofIvy

MHF Helper
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 $$\displaystyle e^{\lambda x}$$

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

so general solution is;

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

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 $$\displaystyle e^{kxi}$$ and $$\displaystyle =ke^{-kxi}$$.

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

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

From that, $$\displaystyle ae^{ix}+ be^{-ix}= a(cos(x)+ i sin(x))+ b(cos(x)- i sin(x))$$$$\displaystyle = acos(x)+ ai sin(x)+ bcos(x)- bi sin(x)$$$$\displaystyle = (a+ b)cos(x)+ i(a- b)sin(x)= A cos(x)+ B sin(x)$$ 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).

• monster