Why Does a Constant Appear Only on One Side in This Differential Equation?

[Marcis Rancans]

I don't understand this first order differential equation:
https://lh5.googleusercontent.com/UUpQF4YjmjJRPvFuzGg2MhpMMMDyi2KFZPCKMKVIXGREc1owvXDzGR0bcA=s600
How is it possible to get an exponent as answer?

 

[SteamKing]

What if Eq. (2) were re-written as follows:

dV/dt = -(E/R) * V

then, separating the variables:

dV/V = -(E/R) dt

Care to take it from here?

 

[Marcis Rancans]

What if Eq. (2) were re-written as follows:

dV/dt = -(E/R) * V

then, separating the variables:

dV/V = -(E/R) dt

Care to take it from here?

I would like to see what comes up next! :)

 

[SteamKing]

I would like to see what comes up next! :)

Why can't you solve this DE now? All you have to do is integrate both sides of the equation.

 

[Marcis Rancans]

Why can't you solve this DE now? All you have to do is integrate both sides of the equation.

I have no idea how to get constant in front of exponent.
https://lh4.googleusercontent.com/5VChRnbmBvEIkRzaZU7vGVBjf1r09ypiNOhcbq4C0LlVdP_3Tqdu_Pvtrg=s600

EDIT: Forgot to add "-C" in exponent function which I got drom integration.

 

[SteamKing]

I have no idea how to get constant in front of exponent.
https://lh4.googleusercontent.com/5VChRnbmBvEIkRzaZU7vGVBjf1r09ypiNOhcbq4C0LlVdP_3Tqdu_Pvtrg=s600

EDIT: Forgot to add "-C" in exponent function which I got drom integration.

If you have ln V = Mess, what do you do to the LHS of the equation to get V? i.e., how is the natural log of a number related to the constant e?

 

[Marcis Rancans]

If you have ln V = Mess, what do you do to the LHS of the equation to get V? i.e., how is the natural log of a number related to the constant e?

Isn't that correct what I wrote? V is e^(MESS).

 

[SteamKing]

Isn't that correct what I wrote? V is e^(MESS).

It is. I'm sorry for not fully recognizing that.

However, when you did the integration to obtain ln V, you forgot to include the constant of integration on the RHS, thus:

∫ dV/V = ∫ (-E/R) dt

ln V = (-E/R)*t + C

Let's say at t = 0, V = V₀, then

ln V₀ = C, so

ln V = (-E/R)*t + ln V₀

exponentiating both sides gives:

e^{ln V} = e^{[(-E/R)*t + ln V₀]}

which can be simplified:

V = e^(-E/R)*t * e^{ln V₀} {using the law of exponents}

V = V₀ * e^(-E/R)*t

if T = R/E, then

V = V₀ * e^{(-t / T)}

 

[Marcis Rancans]

It is. I'm sorry for not fully recognizing that.

However, when you did the integration to obtain ln V, you forgot to include the constant of integration on the RHS, thus:

∫ dV/V = ∫ (-E/R) dt

ln V = (-E/R)*t + C

Let's say at t = 0, V = V₀, then

ln V₀ = C, so

ln V = (-E/R)*t + ln V₀

exponentiating both sides gives:

e^{ln V} = e^{[(-E/R)*t + ln V₀]}

which can be simplified:

V = e^(-E/R)*t * e^{ln V₀} {using the law of exponents}

V = V₀ * e^(-E/R)*t

if T = R/E, then

V = V₀ * e^{(-t / T)}

Thanks for clear explanation! Only thing I don't understand why constant appears only on one side of equation?

 

[SteamKing]

Thanks for clear explanation! Only thing I don't understand why constant appears only on one side of equation?

Technically, integrating both sides of the equation will result in a constant of integration for each integral:

∫ dV/V = ∫ (-E/R) dt

ln V + C₁ = (-E/R) * t + C₂

The two separate constants of integration can be combined into one constant:

ln V = (-E/R) * t + C₂ - C₁ = (-E/R) * t + C, where C = C₂ - C₁

and the solution proceeds as described in Post #8.