# Homework Help: Differential equuation with boundary conditions

1. Aug 24, 2010

### ookt2c

1. The problem statement, all variables and given/known data
d^2T/dx^2+S/K=0 Boundary Conditions T=Tsub1 @ x=0
and T=Tsub2 @ x=L

2. Relevant equations

3. The attempt at a solution

d^2T/dx^2 = -(S/K) <--- intergrate to get
dT=-(S/K)dx+ C1 <--- intergrate to get
T=(-S/K)x+c1+c2
apply both boundary conditions to get
Tsub1=c1+c2

Not sure if i doing it right and if i am i dont know how to get c1 and c2
Tsub2=-(S/K)*L +c1+c2

2. Aug 24, 2010

### gabbagabbahey

You mean "integrate", right And you aren't integrating to get to here, you are simply rearranging your DE algebraically.

No,

$$\frac{d^2T}{dx^2}=-\frac{S}{K} \implies \frac{dT}{dx}=\int -\frac{S}{K}dx= -\frac{S}{K}x+C_1$$

You will need to integrate once more to get $T$

3. Aug 24, 2010

### ookt2c

I meant integrate that equation to get to the next line.

And after I integrate I get: T=-(S/K)* (x^2)/2+C1+C2

Now I just plug in the boudary conditions and solve the system of equations for c1 and c2 correct?

4. Aug 24, 2010

### gabbagabbahey

$$\int(-\frac{S}{K}x+C_1)dx\neq -\frac{S}{K}x^2+C_1 +C_2$$

You're missing something.

5. Aug 24, 2010

### ookt2c

=-S/K* x^2/2+C1x+C2

Also how do I make my equations appear like yours?

6. Aug 24, 2010

### gabbagabbahey

Good, now use your boundary conditions to find C1 and C2

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