Determine p(z) by solving dp/dz = -(1/λ)(p)
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Determine p(z) by solving the differential equation
dp/dz = -1/λ*p
where λ is a constant, and find the particular solution that satisfies the
initial condition p(0) = P , where P is a constant.
I've assumed that using seperation of variables method is suitable as the RHS is a multiple of two functions, is this correct and if so have i approached it in the right way?
1/p dp = λ^1/p dz
Integrate both sides
ln|p|= 1/λ*ln|p|+C
differential-equations initial-value-problems
asked Nov 16 at 14:05
RocketKangaroo
163
add a comment |
up vote
0
down vote
favorite
Determine p(z) by solving the differential equation
dp/dz = -1/λ*p
where λ is a constant, and find the particular solution that satisfies the
initial condition p(0) = P , where P is a constant.
I've assumed that using seperation of variables method is suitable as the RHS is a multiple of two functions, is this correct and if so have i approached it in the right way?
1/p dp = λ^1/p dz
Integrate both sides
ln|p|= 1/λ*ln|p|+C
differential-equations initial-value-problems
asked Nov 16 at 14:05
RocketKangaroo
163
Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Determine p(z) by solving the differential equation
dp/dz = -1/λ*p
where λ is a constant, and find the particular solution that satisfies the
initial condition p(0) = P , where P is a constant.
I've assumed that using seperation of variables method is suitable as the RHS is a multiple of two functions, is this correct and if so have i approached it in the right way?
1/p dp = λ^1/p dz
Integrate both sides
ln|p|= 1/λ*ln|p|+C
differential-equations initial-value-problems
asked Nov 16 at 14:05
RocketKangaroo
163
Determine p(z) by solving the differential equation
dp/dz = -1/λ*p
where λ is a constant, and find the particular solution that satisfies the
initial condition p(0) = P , where P is a constant.
I've assumed that using seperation of variables method is suitable as the RHS is a multiple of two functions, is this correct and if so have i approached it in the right way?
1/p dp = λ^1/p dz
Integrate both sides
ln|p|= 1/λ*ln|p|+C
differential-equations initial-value-problems
differential-equations initial-value-problems
asked Nov 16 at 14:05
RocketKangaroo
163
asked Nov 16 at 14:05
RocketKangaroo
163
edited Nov 16 at 14:22
asked Nov 16 at 14:05
RocketKangaroo
163
asked Nov 16 at 14:05
RocketKangaroo
163
asked Nov 16 at 14:05
RocketKangaroo
163
163
Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20
add a comment |
Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20
Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20
add a comment |
1 Answer
1
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0
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Assuming that $p$ is in the numerator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1lambda p \
frac{dp}p&=-frac{dz}lambda \
ln p &=-frac zlambda + C \
p(z) &= p(0)e^{-frac zlambda}
end{align}
Assuming that $p$ is in the denominator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1{lambda p} \
pdp&=-frac{dz}lambda \
frac12 p^2 &=-frac zlambda + C \
p(z) &= sqrt{p(0)^2-frac{2z}lambda}
end{align}
answered Nov 16 at 14:38
Danijel
776417
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Assuming that $p$ is in the numerator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1lambda p \
frac{dp}p&=-frac{dz}lambda \
ln p &=-frac zlambda + C \
p(z) &= p(0)e^{-frac zlambda}
end{align}
Assuming that $p$ is in the denominator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1{lambda p} \
pdp&=-frac{dz}lambda \
frac12 p^2 &=-frac zlambda + C \
p(z) &= sqrt{p(0)^2-frac{2z}lambda}
end{align}
answered Nov 16 at 14:38
Danijel
776417
add a comment |
up vote
0
down vote
Assuming that $p$ is in the numerator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1lambda p \
frac{dp}p&=-frac{dz}lambda \
ln p &=-frac zlambda + C \
p(z) &= p(0)e^{-frac zlambda}
end{align}
Assuming that $p$ is in the denominator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1{lambda p} \
pdp&=-frac{dz}lambda \
frac12 p^2 &=-frac zlambda + C \
p(z) &= sqrt{p(0)^2-frac{2z}lambda}
end{align}
answered Nov 16 at 14:38
Danijel
776417
add a comment |
up vote
0
down vote
up vote
0
down vote
Assuming that $p$ is in the numerator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1lambda p \
frac{dp}p&=-frac{dz}lambda \
ln p &=-frac zlambda + C \
p(z) &= p(0)e^{-frac zlambda}
end{align}
Assuming that $p$ is in the denominator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1{lambda p} \
pdp&=-frac{dz}lambda \
frac12 p^2 &=-frac zlambda + C \
p(z) &= sqrt{p(0)^2-frac{2z}lambda}
end{align}
answered Nov 16 at 14:38
Danijel
776417
Assuming that $p$ is in the numerator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1lambda p \
frac{dp}p&=-frac{dz}lambda \
ln p &=-frac zlambda + C \
p(z) &= p(0)e^{-frac zlambda}
end{align}
Assuming that $p$ is in the denominator on the RHS:
begin{align}
frac{dp}{dz} &= -frac1{lambda p} \
pdp&=-frac{dz}lambda \
frac12 p^2 &=-frac zlambda + C \
p(z) &= sqrt{p(0)^2-frac{2z}lambda}
end{align}
answered Nov 16 at 14:38
Danijel
776417
answered Nov 16 at 14:38
Danijel
776417
answered Nov 16 at 14:38
Danijel
776417
answered Nov 16 at 14:38
Danijel
776417
776417
add a comment |
add a comment |
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Please use Latex/MathJax notation. Do you now which $f$ satisfies $f' = cf$?
– Stockfish
Nov 16 at 14:11
It is hard to parse your expression. Do you mean $-frac 1{lambda p}$ or $-frac 1{lambda}p$?
– lulu
Nov 16 at 14:12
Sorry guys, struggled to figure out how to put the greek letter in. -(1/λ)(p)
– RocketKangaroo
Nov 16 at 14:20