Solving differential eqution in a specific form
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for the following differential equation
$$frac{dG}{dt}=frac{1}{2}lambda G^{2}+rho G-1$$
$lambda$ and $rho$ are constant, G is a function of t
This is then expressed in the form
$$int_{0}^{G}frac{1}{(G'-a)(G'+b)}dG'=frac{1}{2}lambdaint_{T}^{t}dt$$
where
$$b,a=frac{pm rho +sqrt{rho ^ 2 + 2lambda}}{lambda}$$
I need to show that
$$G(t;T) = frac{2(e^{psi (T-t)}-1)}{(rho+psi)(e^{psi(T-t)}-1)+2psi}$$
where
$$psi = sqrt{rho^2+2lambda}$$
many thanks!
calculus differential-equations
edited Nov 26 at 5:09
gt6989b
32.7k22451
asked Nov 26 at 4:59
2tin
82
add a comment |
up vote
1
down vote
favorite
for the following differential equation
$$frac{dG}{dt}=frac{1}{2}lambda G^{2}+rho G-1$$
$lambda$ and $rho$ are constant, G is a function of t
This is then expressed in the form
$$int_{0}^{G}frac{1}{(G'-a)(G'+b)}dG'=frac{1}{2}lambdaint_{T}^{t}dt$$
where
$$b,a=frac{pm rho +sqrt{rho ^ 2 + 2lambda}}{lambda}$$
I need to show that
$$G(t;T) = frac{2(e^{psi (T-t)}-1)}{(rho+psi)(e^{psi(T-t)}-1)+2psi}$$
where
$$psi = sqrt{rho^2+2lambda}$$
many thanks!
calculus differential-equations
edited Nov 26 at 5:09
gt6989b
32.7k22451
asked Nov 26 at 4:59
2tin
82
Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
thanks for hint
– 2tin
Nov 28 at 5:40
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
for the following differential equation
$$frac{dG}{dt}=frac{1}{2}lambda G^{2}+rho G-1$$
$lambda$ and $rho$ are constant, G is a function of t
This is then expressed in the form
$$int_{0}^{G}frac{1}{(G'-a)(G'+b)}dG'=frac{1}{2}lambdaint_{T}^{t}dt$$
where
$$b,a=frac{pm rho +sqrt{rho ^ 2 + 2lambda}}{lambda}$$
I need to show that
$$G(t;T) = frac{2(e^{psi (T-t)}-1)}{(rho+psi)(e^{psi(T-t)}-1)+2psi}$$
where
$$psi = sqrt{rho^2+2lambda}$$
many thanks!
calculus differential-equations
edited Nov 26 at 5:09
gt6989b
32.7k22451
asked Nov 26 at 4:59
2tin
82
for the following differential equation
$$frac{dG}{dt}=frac{1}{2}lambda G^{2}+rho G-1$$
$lambda$ and $rho$ are constant, G is a function of t
This is then expressed in the form
$$int_{0}^{G}frac{1}{(G'-a)(G'+b)}dG'=frac{1}{2}lambdaint_{T}^{t}dt$$
where
$$b,a=frac{pm rho +sqrt{rho ^ 2 + 2lambda}}{lambda}$$
I need to show that
$$G(t;T) = frac{2(e^{psi (T-t)}-1)}{(rho+psi)(e^{psi(T-t)}-1)+2psi}$$
where
$$psi = sqrt{rho^2+2lambda}$$
many thanks!
calculus differential-equations
calculus differential-equations
edited Nov 26 at 5:09
gt6989b
32.7k22451
asked Nov 26 at 4:59
2tin
82
edited Nov 26 at 5:09
gt6989b
32.7k22451
asked Nov 26 at 4:59
2tin
82
edited Nov 26 at 5:09
gt6989b
32.7k22451
edited Nov 26 at 5:09
gt6989b
32.7k22451
edited Nov 26 at 5:09
gt6989b
32.7k22451
32.7k22451
asked Nov 26 at 4:59
2tin
82
asked Nov 26 at 4:59
2tin
82
asked Nov 26 at 4:59
2tin
82
82
Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
thanks for hint
– 2tin
Nov 28 at 5:40
add a comment |
Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
thanks for hint
– 2tin
Nov 28 at 5:40
Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
thanks for hint
– 2tin
Nov 28 at 5:40
thanks for hint
– 2tin
Nov 28 at 5:40
add a comment |
1 Answer
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I'm going to assume that your initial condition is $G(T)=0$. This restricts the domain of integration to $G in (-b,a)$
Using partial fractions
$$ frac{1}{(a-G')(b+G')} = frac{1}{a+b}left(frac{1}{a-G'}+frac{1}{b+G'}right) $$
where $a+b= frac{2psi}{lambda}$. Integrating the above gives
$$ frac{lambda}{2psi}big[ln(b+G)-ln(a-G) - ln b + ln abig] = frac{lambda}{2}(T-t) $$
or
$$ frac{a(b+G)}{b(a-G)} = e^{psi(T-t)} $$
Do some algebra to get
$$ G = frac{abbig(e^{psi(T-t)}-1big)}{be^{psi(T-t)}+a} $$
You have $ab = frac{2}{lambda}$ so this simplifies to
$$ G = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)e^{psi(T-t)}+(psi-rho)} = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)big(e^{psi(T-t)}-1big)+2psi} $$
answered Nov 26 at 13:22
Dylan
12k31026
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I'm going to assume that your initial condition is $G(T)=0$. This restricts the domain of integration to $G in (-b,a)$
Using partial fractions
$$ frac{1}{(a-G')(b+G')} = frac{1}{a+b}left(frac{1}{a-G'}+frac{1}{b+G'}right) $$
where $a+b= frac{2psi}{lambda}$. Integrating the above gives
$$ frac{lambda}{2psi}big[ln(b+G)-ln(a-G) - ln b + ln abig] = frac{lambda}{2}(T-t) $$
or
$$ frac{a(b+G)}{b(a-G)} = e^{psi(T-t)} $$
Do some algebra to get
$$ G = frac{abbig(e^{psi(T-t)}-1big)}{be^{psi(T-t)}+a} $$
You have $ab = frac{2}{lambda}$ so this simplifies to
$$ G = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)e^{psi(T-t)}+(psi-rho)} = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)big(e^{psi(T-t)}-1big)+2psi} $$
answered Nov 26 at 13:22
Dylan
12k31026
add a comment |
up vote
0
down vote
accepted
I'm going to assume that your initial condition is $G(T)=0$. This restricts the domain of integration to $G in (-b,a)$
Using partial fractions
$$ frac{1}{(a-G')(b+G')} = frac{1}{a+b}left(frac{1}{a-G'}+frac{1}{b+G'}right) $$
where $a+b= frac{2psi}{lambda}$. Integrating the above gives
$$ frac{lambda}{2psi}big[ln(b+G)-ln(a-G) - ln b + ln abig] = frac{lambda}{2}(T-t) $$
or
$$ frac{a(b+G)}{b(a-G)} = e^{psi(T-t)} $$
Do some algebra to get
$$ G = frac{abbig(e^{psi(T-t)}-1big)}{be^{psi(T-t)}+a} $$
You have $ab = frac{2}{lambda}$ so this simplifies to
$$ G = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)e^{psi(T-t)}+(psi-rho)} = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)big(e^{psi(T-t)}-1big)+2psi} $$
answered Nov 26 at 13:22
Dylan
12k31026
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I'm going to assume that your initial condition is $G(T)=0$. This restricts the domain of integration to $G in (-b,a)$
Using partial fractions
$$ frac{1}{(a-G')(b+G')} = frac{1}{a+b}left(frac{1}{a-G'}+frac{1}{b+G'}right) $$
where $a+b= frac{2psi}{lambda}$. Integrating the above gives
$$ frac{lambda}{2psi}big[ln(b+G)-ln(a-G) - ln b + ln abig] = frac{lambda}{2}(T-t) $$
or
$$ frac{a(b+G)}{b(a-G)} = e^{psi(T-t)} $$
Do some algebra to get
$$ G = frac{abbig(e^{psi(T-t)}-1big)}{be^{psi(T-t)}+a} $$
You have $ab = frac{2}{lambda}$ so this simplifies to
$$ G = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)e^{psi(T-t)}+(psi-rho)} = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)big(e^{psi(T-t)}-1big)+2psi} $$
answered Nov 26 at 13:22
Dylan
12k31026
I'm going to assume that your initial condition is $G(T)=0$. This restricts the domain of integration to $G in (-b,a)$
Using partial fractions
$$ frac{1}{(a-G')(b+G')} = frac{1}{a+b}left(frac{1}{a-G'}+frac{1}{b+G'}right) $$
where $a+b= frac{2psi}{lambda}$. Integrating the above gives
$$ frac{lambda}{2psi}big[ln(b+G)-ln(a-G) - ln b + ln abig] = frac{lambda}{2}(T-t) $$
or
$$ frac{a(b+G)}{b(a-G)} = e^{psi(T-t)} $$
Do some algebra to get
$$ G = frac{abbig(e^{psi(T-t)}-1big)}{be^{psi(T-t)}+a} $$
You have $ab = frac{2}{lambda}$ so this simplifies to
$$ G = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)e^{psi(T-t)}+(psi-rho)} = frac{2big(e^{psi(T-t)}-1big)}{(psi+rho)big(e^{psi(T-t)}-1big)+2psi} $$
answered Nov 26 at 13:22
Dylan
12k31026
edited Nov 26 at 13:39
answered Nov 26 at 13:22
Dylan
12k31026
answered Nov 26 at 13:22
Dylan
12k31026
answered Nov 26 at 13:22
Dylan
12k31026
12k31026
add a comment |
add a comment |
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Just integrate using partial fractions. Here you'll find similar problems: google.com/…
– Hans Lundmark
Nov 26 at 5:59
thanks for hint
– 2tin
Nov 28 at 5:40