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1.

Let c'(t) = f(c) = [itex]\frac{(kr + P - c(t)r)}{V}[/itex]. Determine the singular solution of Vc'(t)

2.

k, P, r, and V are all constant

I'm not quite sure if this is the correct way to find the singular solution but I used this equation:

Integrating Factor Method:

To solve a linear DE

y' + x(t)y = f(t)

where x and f are continuous on the domain I:

u(t) = e[itex]^\int{x(t) dt}[/itex]

u(t)[y' + x(t)y] = f(t)u(t)

[itex]\frac{d}{dt}[/itex][u(t)y(t)] = f(t)u(t)

u(t)y(t) = [itex]\int{f(t)u(t )dt}[/itex] + j

where j is a constant

y = ([itex]\frac{\int{f(t)u(t) dt} + j)}{u(t)}[/itex]

3.

f'(c) = -[itex]\frac{c'(t)r}{V}[/itex]

f'(c) = -[itex]\frac{f(c)r}{V}[/itex]

f'(c) + [itex]\frac{f(c)r}{V}[/itex] = 0

let x = [itex]\frac{r}{V}[/itex]

u = e[itex]^{\int{xdt}}[/itex]

u = e[itex]^{\frac{d}{dt}}[/itex]

[itex]\frac{d}{dt}[/itex][e[itex]^{\frac{rt}{V}}[/itex]f(c)] = 0

e[itex]^{\frac{rt}{V}}[/itex]f(c) = j

f(c) = [itex]\frac{j}{e^\frac{rt}{V}}[/itex]

There is no singular solution

Now that can't be right can it? I'm completely confused and now I'm second guessing myself everywhere. Any help?

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# Singular solution for a differential equation?

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