# Simple Differential problem (help)

## Main Question or Discussion Point

Hi,

I am working out a heat transfer problem but i've to solve the Differential equation in order to keep going on but it's been a long time since i did any Differential. your help will be appreciated.

d/dx[(k+aT)dT/dx]=0

i need thins in terms of T(x)

my boundary conditions T(x=0)=To
T(x=L)=TL

my work

d/dx[(k+aT)dT/dx]=0
(k+aT)dT=C1dx

KT+(a/2)T^2=C1x+C2

rearranging

T^2+(2/a)*k*T=(2/a)*(C1x+C2)

lets call (2/a)=z

T^2+z*k*T=z*(C1x+C2)

i could not get further from here due to lack of my Deferential equation knowledge. I really need to learn this as i'm expecting to see a lot more of this later and i actually took differential long time ago. any help would be appreciated

Mark44
Mentor
Hi,

I am working out a heat transfer problem but i've to solve the Differential equation in order to keep going on but it's been a long time since i did any Differential. your help will be appreciated.

d/dx[(k+aT)dT/dx]=0
Since the derivative, with respect to x, of the quantity in brackets is zero, then the stuff in brackets must be a constant.

I.e., (k + aT)dT/dx = C

This equation is separable, so put all the terms involving T and dT on one side, and all the terms involving x and dx on the other side, and integrate.
i need thins in terms of T(x)

my boundary conditions T(x=0)=To
T(x=L)=TL

my work

d/dx[(k+aT)dT/dx]=0
(k+aT)dT=C1dx

KT+(a/2)T^2=C1x+C2

rearranging

T^2+(2/a)*k*T=(2/a)*(C1x+C2)
This equation is quadratic in T, so use the quadratic formula to solve for T.

Write the equation in standard form, as
T^2+(2k/a)*T - (2/a)(C1x+C2) = 0

lets call (2/a)=z

T^2+z*k*T=z*(C1x+C2)

i could not get further from here due to lack of my Deferential equation knowledge. I really need to learn this as i'm expecting to see a lot more of this later and i actually took differential long time ago. any help would be appreciated
Where you got stumped doesn't have anything to do with differential equations - this is pretty much plain algebra.

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