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Heat transfer between two fluids through a sandwiched solid (coupled problem):
Two fluids ${\displaystyle (t_h,t_c)}$ flow opposite to each other on either side of a solid (T), while exchanging heat among themselves. In such a scenario, the conduction in the solid is governed by:
${\displaystyle x\in [0,1],y\in [0,1]}$
1) $\frac{dT}{d{x}^2}+\mu b_h(t_h-T)-\nu b_c(T-t_c)=0$
with boundary condition as ${\displaystyle T^{\prime }\left(0\right)=T^{\prime }\left(1\right)=0}$
The fluids are governed by the following equations:
2) $\frac{d{t}_h}{dx}+b_h(t_h-T)=0$
3) $\frac{d{t}_c}{dx}+b_c(T-t_c)=0$
The hot fluid initiates at ${\displaystyle x=0}$ and the cold fluid starts from ${\displaystyle x=1}$.
The boundary conditions are ${\displaystyle t_h(x=0)=1}$ and ${\displaystyle t_c(x=1)=0}$
Equation (1), (2) and (3) form a coupled system of ordinary differential equations.
It is pretty evident that using (2) and (3), Equation (1) can be re-written as:
4) $\frac{d^{2}T}{d{x}^2}-\mu \frac{d{t}_h}{dx}+\nu \frac{d{t}_c}{dx}=0$
However, I have not been able to proceed further.
Some parameter values are
${\displaystyle b_c=12.38,b_h=25.32,\mu =1.143,\nu =1,\kappa =2.16}$

How to solve homogenous differential equation ${(x+y)}^2 dx=xy dy$?
First i find value of $\frac{dy}{dx}=\frac{{(x+y)}^2}{x}y \left(i\right)$
let ${\displaystyle y=vx}$
differentiating both sides: $\frac{dy}{dx}=v+x d\frac{v}{dx} \left(ii\right)$
I tried to solve using equation (i) and (ii) but I am stuck.
${(x+y)}^2 dx=xy dy$

If n = 20 and p = 0.70, then P(X < 2)  is

1. f(x) = 12 xe6x

How a solving a nonlinear PDE with Self-similar Solution?
I want to solve the problem :
$\frac{\mathrm{\partial }u}{\mathrm{\partial }t}={\left|\mathrm{\nabla }u\right|}^{-2p}(\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{\xi }^2}+(1-p\left)\frac{{\mathrm{\partial }}^{2}u}{\mathrm{\partial }{\eta }^2}\right)$
We search for a self-similarity solution, the general form of which is as follows
${\displaystyle u(x,y,t)=f\left(\xi \right),\text{ with }\xi =d\frac{{(x^2+y^2)}^n}{a\left(t\right)}}$
from which we obtain
${\displaystyle \alpha \xi =(1-p){\left(2n\right)}^{-2p+2}((\frac{1}{2n(1-p)}+\frac{2n-1}{2n}){\left(d\frac{df}{d\xi }\right)}^{-2p}+\xi {\left(d\frac{df}{d\xi }\right)}^{-2p-1}d\frac{d^{2}f}{d{\xi }^2})}$
Now, I am very confused on how to solve the above equation and find the exact solution of ${\displaystyle f\left(\xi \right)}$, thereby finding the exact solution of $u(x,y,t)$. So my question is, does anyone know how to solve this ordinary differential equation?

In 2008, the population of a district was 36,700. With a continuous annual growth rate of approximately 2%, what will the population be in 2028 according to the exponential growth function?

Solving a systemof eqns by given initial conditions
${\displaystyle x^{\prime }\left(t\right)=2x+8y}$
${\displaystyle y^{\prime }\left(t\right)=-x-2y}$
Which has the I.C. ${\displaystyle X\left(0\right)=(6,-2)}$. So I take that it means this:
${\displaystyle 6=2x+8y}$
${\displaystyle -2=-x-2y}$
and then it is solved as any other system? We get eigenvector ${\displaystyle (1,\frac{1}{2})}$, but since the eigenvalues of the system are two, ${\displaystyle \pm 2i}$, we have to find the second generalized eigenvector.
$\left(\begin{array}{cc}2-2i& 8\\ -1& -2-2i\end{array}\right)\left(\begin{array}{c}1\\ \frac{1}{2}\end{array}\right)=\left(\begin{array}{c}6-2i\\ -2-i\end{array}\right)$
So the second vector would be ${\displaystyle (6-2i,-2-i)}$. Using the general solutions for the system, we get
$y\left(t\right)=e^{2it}(1,\frac{1}{2})+e^{-2it}(6-2i,-2-i)$
Would this be correct, or did I misinterpret the initial conditions?