Using an integrating factor to solve an ODE dcdt=1−c−ac

Name: Using an integrating factor to solve an ODE dcdt=1−c−ac
Uploaded: 2022-02-23T17:22:57+00:00
Description: Using an integrating factor to solve an ODE dcdt=1−c−ac

dcdt+(1+a)c=1 multiply by integrating factor e(1+a)t: e(1+a)tdcdt+(1+a)e(1+a)tc=e(1+a)t ddt(e(1+a)tc)=e(1+a)t e(1+a)tc=∫e(1+a)tdt=e(1+a)t1+a+A c=11+a+Ae−(1+a)t for some constant A. Presumably there is an initial condition (that R.M. has not told us) to let the lecturer determine the constant A

t=∫dc1−(a+1)c =−1a+1log⁡(1−(a+1)c)+t0 (1) Solving (1) yields c=1a+1(1−e−(a+1)(t−t0)) =1a+1−(1a+1−c0)e−(a+1)t (2) c=1a+1 is a particular solution of (2) with c0=1a+1(t0=−∞)

Because of the nature of this equation it is separable so an integrating factor is not strictly necessary. but because you requested it be done in that method: first consider the form of the equation for an integrating factor:A differential equation of the form y′+P(x)y=Q(x) can be solved byy=∫μQ(x)dxμ+kμ where k is an arbitrary constant and μ is the integrating factor (μ=e∫P(x)dx)Let's fit your equation into the form needed for the method of integrating factors (In your case, y is c ):c′+c(1+a)=1so you should obtain for an integrating factor of μ=e∫(1+a)dt=et(1+a)so c=1et(1+a)∫(et(1+a))dt+ket(1+a)if you integrated correctly, you should get ∫(et(1+a))dt=11+aet(1+a) which you will find nicely cancels with the integrating factor, μ which is already in the denominator. So your final answer should bec=11+a+ket(1+a)=11+a+ke−t(1+a)

dcdt+(1+a)c=1 multiply by integrating factor e(1+a)t: e(1+a)tdcdt+(1+a)e(1+a)tc=e(1+a)t ddt(e(1+a)tc)=e(1+a)t e(1+a)tc=∫e(1+a)tdt=e(1+a)t1+a+A c=11+a+Ae−(1+a)t for some constant A. Presumably there is an initial condition (that R.M. has not told us) to let the lecturer determine the constant A

t=∫dc1−(a+1)c =−1a+1log⁡(1−(a+1)c)+t0 (1) Solving (1) yields c=1a+1(1−e−(a+1)(t−t0)) =1a+1−(1a+1−c0)e−(a+1)t (2) c=1a+1 is a particular solution of (2) with c0=1a+1(t0=−∞)

Because of the nature of this equation it is separable so an integrating factor is not strictly necessary. but because you requested it be done in that method: first consider the form of the equation for an integrating factor:A differential equation of the form y′+P(x)y=Q(x) can be solved byy=∫μQ(x)dxμ+kμ where k is an arbitrary constant and μ is the integrating factor (μ=e∫P(x)dx)Let's fit your equation into the form needed for the method of integrating factors (In your case, y is c ):c′+c(1+a)=1so you should obtain for an integrating factor of μ=e∫(1+a)dt=et(1+a)so c=1et(1+a)∫(et(1+a))dt+ket(1+a)if you integrated correctly, you should get ∫(et(1+a))dt=11+aet(1+a) which you will find nicely cancels with the integrating factor, μ which is already in the denominator. So your final answer should bec=11+a+ket(1+a)=11+a+ke−t(1+a)

Key to "Using an integrating factor to solve an ODE..."

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Answered question

2022-01-21

Using an integrating factor to solve an ODE

\frac{d c}{d t} = 1 - c - a c

Answer & Explanation

MoxboasteBots5h

Beginner2022-01-21Added 35 answers

\frac{d c}{d t} + (1 + a) c = 1

multiply by integrating factor

e^{(1 + a) t}

e^{(1 + a) t} \frac{d c}{d t} + (1 + a) e^{(1 + a) t} c = e^{(1 + a) t}

\frac{d}{d t} (e^{(1 + a) t} c) = e^{(1 + a) t}

e^{(1 + a) t} c = \int e^{(1 + a) t} d t = \frac{e^{(1 + a) t}}{1 + a} + A

c = \frac{1}{1 + a} + A e^{- (1 + a) t}

for some constant A. Presumably there is an initial condition (that R.M. has not told us) to let the lecturer determine the constant A

abonirali59

Beginner2022-01-22Added 35 answers

t = \int \frac{d c}{1 - (a + 1) c}

= - \frac{1}{a + 1} \log (1 - (a + 1) c) + t_{0}

(1)
Solving (1) yields

c = \frac{1}{a + 1} (1 - e^{- (a + 1) (t - t_{0})})

= \frac{1}{a + 1} - (\frac{1}{a + 1} - c_{0}) e^{- (a + 1) t}

(2)

c = \frac{1}{a + 1}

is a particular solution of (2) with

c_{0} = \frac{1}{a + 1} (t_{0} = - \infty)

RizerMix

Expert2022-01-27Added 656 answers

Because of the nature of this equation it is separable so an integrating factor is not strictly necessary. but because you requested it be done in that method: first consider the form of the equation for an integrating factor:A differential equation of the form

y^{'} + P (x) y = Q (x)

can be solved by

y = \frac{\int μ Q (x) d x}{μ} + \frac{k}{μ}

where k is an arbitrary constant and

μ

is the integrating factor

(μ = e^{\int P (x) d x})

Let's fit your equation into the form needed for the method of integrating factors (In your case, y is c ):

c^{'} + c (1 + a) = 1

so you should obtain for an integrating factor of

μ = e^{\int (1 + a) d t} = e^{t (1 + a)}

c = \frac{1}{e^{t (1 + a)}} \int (e^{t (1 + a)}) d t + \frac{k}{e^{t (1 + a)}}

if you integrated correctly, you should get

\int (e^{t (1 + a)}) d t = \frac{1}{1 + a} e^{t (1 + a)}

which you will find nicely cancels with the integrating factor,

μ

which is already in the denominator. So your final answer should be

c = \frac{1}{1 + a} + \frac{k}{e^{t (1 + a)}} = \frac{1}{1 + a} + k e^{- t (1 + a)}

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