$$abc \frac{dT}{dt} = s+q- k(T-Y)$$
How to get the value of $T$?
My Solution
$$T = Y+ \frac{sq}{k} + \{T(0)-Y- \frac{sq}{k}\} \exp (\frac{-k}{abc})t$$
is it correct ?
High School Math $abc \frac{dT}{dt} = s+q- k(T-Y)$
0
$\begingroup$
ordinary-differential-equations
-
0can you please describe what are those $a,b,c,s,q,k,$ and $Y$? – 2017-01-04
-
0constant numerical value. I need to calculate the numerical value of T for corresponding values of a,b,c,s,q,Y. In my project I can take the numerical values of a,b,c,s,q,Y randomly – 2017-01-04
-
0Since you post an answer, much better if you show some details on how you come up with it. That's a separable DE anyway and easy to solve. – 2017-01-04
-
0Let , R=T-Y , then , dR/dt=dT/dt >> dR/dt= (s+q)/abc - kR/abc >> let , M=(s+q)/abc , and N=k/abc >> so , dR/dt= M-NR – 2017-01-04
1 Answers
0
Considering the equation $$abc \frac{dT}{dt} = s+q- k(T-Y)$$ why not to make life easier setting first $$X=s+q- k(T-Y)$$ In such a case (assuming $k\neq 0$), the equation becomes $$a b c X'(t)+k X(t)=0\implies X=c_1 e^{-\frac{k t}{a b c}}$$ and now, back to $T$ $$T=\frac{q+s-X}{k}+Y$$
-
0what is the value of C1 for overall equation? – 2017-01-04
-
0@Sid. It is fixed using $T(0)=T_0$ – 2017-01-04