Need help solving simple Differential problem (help)

[Jennifer_88]

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.

the equation in heat transfer form is T^2+z*k*T=z(C1*x+C2)

or

d^2y/dx+z*k*dy/dx=z(C1*x+C2)

z & k are constants, the equation need to be solved in terms of y(x)

 

[Delta2]

if you set f=dy/dx then it becomes $$df/dx+zkf=z(c_1x+c_2)$$. You multiply by the integrating factor $$e^{zkx}$$ and get
$$(f(x)e^{zkx})'=ze^{zkx}(c_1x+c_2)$$

and by integrating both sides and solving for f(x) you ll get

$$f(x)=\frac{z\int c_1xe^{zkx}dx+ z\int c_2e^{zkx}dx+ c_3}{e^{zkx}}$$. You just have to compute the integrals which seem easy and get f(x). You then find $$y(x)=\int f(x)dx$$

 

[Jennifer_88]

where did C3 come from ?? thanks for the help

 

[Delta2]

where did C3 come from ?? thanks for the help

It is the integration constant. You can calculate by the initial condition for f(=dy/dx). You ll have another c4 constant from the integration of f to find y.

 

[blue_raver22]

Hi there,

Thank you for reaching out for help with your heat transfer problem. Differential equations can be challenging, but I am happy to assist you in solving this problem.

Firstly, it would be helpful to know the initial conditions or boundary conditions for the problem, as these will be necessary for solving the differential equation. Additionally, it would be useful to know the physical interpretation of the constants z and k in the heat transfer equation.

To solve this differential equation, we can use standard techniques such as separation of variables, integrating factors, or substitution methods. I would recommend starting by rearranging the equation in terms of y(x) and its derivatives, and then applying the chosen method to solve for y(x).

If you need further assistance, please provide more information and I would be happy to guide you through the problem step by step. Good luck!