Integration question - please help

Question

I am studying the heat equation and how to solve it using separation of variables and PDEs. I came across this in my lecture notes:

$$\frac{X''(x)}{X(x)}=k^2 \Longrightarrow X''(x)-k^2X(x) = 0 \Longrightarrow X(x) = Ae^{kx}+Be^{-kx}$$

and

$$\frac{X''(x)}{X(x)} = 0 \Longrightarrow X(x) = A+Bx$$

What I don't understand here is how the results $Ae^{kx} + Be^{kx}$ and $A+Bx$ came about from the ODEs on the left hand side to it.

Any help is hugely appreciated - thank you :)

Answer 1

Integration, indeed:

$$\frac{X''(x)}{X(x)}=0\implies X''(x)=0\implies C=X'(x)=\int X''(x) dx\;,\;C=\text{constant}\implies$$

$$X(x)=\int X'(x)dx=\int Cdx=Cx+D\;,\;\;D=\text{constant}$$

For the other case:

$$\frac{X''(x)}{X(x)}=k^2\implies X''(x)=k^2X(x)\iff \frac{dX'(x)}{dx}=k^2X(x)\implies$$

$$\int dX'(x)=k2\int X(x)dx\iff X'(x)+A=k^2\int X(x)dx$$

and etc.

Answer 2

first the condition of your problem suggests that the separation constant should be zero or negative or positive In case of negative it will be $-\lambda^2$ and the equation becomes $X''+\lambda X=0$. This is a second orser linear DE. suppose that the solution is $X=e^{mx}$ By substitution we get $m^2+\lambda^2=0$ implies $m=-\lambda i$ and $+\lambda i$ This gives $A\cos\lambda x+B\sin \lambda x$
The zero case gives you $X''=0$ by integration twice w.r.t $x$ you get your solution
If you consider the positive case:suppose that the solution is $X=e^{mx}$ By substitution we get $m^2-\lambda^2=0$ implies $m=-\lambda $ and $+\lambda $ This gives $Ae^{\lambda x}+Be^{\sin \lambda x}$