cant get to the same expression as the solution..

[nabliat6]

**broken link removed**
i tried to solve it by myself but
in my expression as you see i get a totally different expression when i do coefficient of exponent then the one in the solution
did i do it correctly
[latex][CR_1+C\alpha \beta R_2]\frac{-\beta }{\tau }e^\frac{-t }{\tau }+Be^\frac{-t }{\tau }=V_0e^\frac{-t }{\tau }[/latex]

the alpha and beta dissapeared
why?

 

[MrAl]

Hi there,


A quick glance makes it look like they are simply lumping R1+alpha*beta*R2 into R as:
R=R1+a*b*R2

where a=alpha and b=beta and they are both constants.

If this is the case then you would get the same answer as they did.

 

[sam77]

Lookup linear first order differential equations
There is a topic that uses the integrating factor to solve this type of equations
Ex
Eq1

dV(t)/dt*k+V(t)=O(t)​

Eq2

dV(t)/dt+1/k*V(t)=O​

We use an integrating factor

e^(∫1/k dt )=e^(t/k)
​

We multiply eq2 by the integrating factor
Eq3

e^(t/k)*dV(t)/dt+e^(t/k)*1/k*V(t)=e^(t/k)*O
​

We notice that the left hand side of equation 3 is the derivative of a product, namely (Calculus 1)

d/dt (e^(t/k)*V(t) )=e^(t/k)*dV(t)/dt+e^(t/k)*1/k*V(t)
​

We make the substitution in eq3

d/dt (e^(t/k)*V(t) )=e^(t/k)*O
​

Integrate both sides

(e^(t/k)*V(t) )=∫〖e^(t/k)*O dt 〗

e^(t/k)*V(t)=e^(t/k)*O+c

V(t)=(〖(e〗^(t/k)*O+c))/e^(t/k) =O+c/e^(t/k)
​

Initial condition not applied
I also ran the Differential equation (DE) in Maple and the answer is correct. You also have to consider that this is not the only solution to the DE. ( review ODE)
hope it helped.
Can Math be written on this forum?