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










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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















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










share|cite|improve this question
























  • 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













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










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













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edited Nov 16 at 14:22

























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


















  • 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










1 Answer
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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}






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    1 Answer
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    1 Answer
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    active

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    active

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    up vote
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    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}






    share|cite|improve this answer

























      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}






      share|cite|improve this answer























        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}






        share|cite|improve this answer












        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}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 16 at 14:38









        Danijel

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        776417






























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