solve differential equation $frac{dP}{dt} = kPcos^{2}(rt-Theta)$












1














I'm asked to solve the following differential equation.



$$frac{dP}{dt} = kP cos^{2}(rt-theta)$$



$P(0) = P_{0} = 9$ for $k = 0.07, r = 0.49, theta = 7.$



I've only done simple linear and separable differential equations up until this point, so I'm not sure how to approach this one. Any pointers or solutions are much appreciated.






calculus differential-equations







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  • 3




    Do you not think it's linear? Do you not think it's separable?
    – David
    Dec 10 '18 at 2:55
















1














I'm asked to solve the following differential equation.



$$frac{dP}{dt} = kP cos^{2}(rt-theta)$$



$P(0) = P_{0} = 9$ for $k = 0.07, r = 0.49, theta = 7.$



I've only done simple linear and separable differential equations up until this point, so I'm not sure how to approach this one. Any pointers or solutions are much appreciated.






calculus differential-equations







share|cite|improve this question




















  • 3




    Do you not think it's linear? Do you not think it's separable?
    – David
    Dec 10 '18 at 2:55














1












1








1


1





I'm asked to solve the following differential equation.



$$frac{dP}{dt} = kP cos^{2}(rt-theta)$$



$P(0) = P_{0} = 9$ for $k = 0.07, r = 0.49, theta = 7.$



I've only done simple linear and separable differential equations up until this point, so I'm not sure how to approach this one. Any pointers or solutions are much appreciated.






calculus differential-equations







share|cite|improve this question















I'm asked to solve the following differential equation.



$$frac{dP}{dt} = kP cos^{2}(rt-theta)$$



$P(0) = P_{0} = 9$ for $k = 0.07, r = 0.49, theta = 7.$



I've only done simple linear and separable differential equations up until this point, so I'm not sure how to approach this one. Any pointers or solutions are much appreciated.






calculus differential-equations




calculus differential-equations






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edited Dec 10 '18 at 3:13









Mattos

2,73021321




2,73021321










asked Dec 10 '18 at 2:50









BOBBY SHMURDA

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102








  • 3




    Do you not think it's linear? Do you not think it's separable?
    – David
    Dec 10 '18 at 2:55














  • 3




    Do you not think it's linear? Do you not think it's separable?
    – David
    Dec 10 '18 at 2:55








3




3




Do you not think it's linear? Do you not think it's separable?
– David
Dec 10 '18 at 2:55




Do you not think it's linear? Do you not think it's separable?
– David
Dec 10 '18 at 2:55










1 Answer
1






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oldest

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1














The equation



$dfrac{dP}{dt} = kP cos^2 (rt - Theta) tag 1$



is in fact of the variables-seperable type, to wit:



If $P(t') = 0$ for any $t' in Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t in Bbb R$, since the zero solution satisfies $P(t') = 0$. Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$. Therefore, for any solution which does not vanish identically we may write



$dfrac{dln P}{dt} = dfrac{1}{P}dfrac{dP}{dt} = k cos^2 (rt - Theta); tag 2$



we integrate 'twixt $t_0$ and $t$:



$ln dfrac{P(t)}{P(t_0)} = ln P(t) - ln P(t_0) = k displaystyle int_{t_0}^t cos^2(rt - Theta) ; dt; tag 3$



we have in general (from a table of integrals):



$displaystyle int cos^2 ax ; dx = dfrac{x}{2} + dfrac{sin 2ax}{4a}; tag 4$



setting



$alpha = dfrac{Theta}{r}, tag 5$



we write



$rt - Theta = r(t - alpha), tag 6$



and



$displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = int_{t_0}^t cos^2 r(t - alpha) ; dt = left ( dfrac{t - alpha}{2} + dfrac{sin 2r(t - alpha)}{4r} right vert_{t_0}^t$
$= dfrac{t - t_0}{2} + dfrac{sin 2r(t - alpha) - sin 2r(t_0 - alpha)}{4r}$
$= left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r}right ) - left ( dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r} right ); tag 7$



setting



$beta(t_0) = dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r}, tag 8$



we write



$displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} - beta(t_0); tag 9$



returning to (3),



$ln dfrac{P(t)}{P(t_0)} = k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0), tag{10}$



or



$P(t) = P(t_0) exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0) right )$
$= P(t_0) e^{ - kbeta(t_0) } exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right )right ); tag{11}$



there is not much to be gained at this point by furhter re-arrangement of (11). The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.






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    The equation



    $dfrac{dP}{dt} = kP cos^2 (rt - Theta) tag 1$



    is in fact of the variables-seperable type, to wit:



    If $P(t') = 0$ for any $t' in Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t in Bbb R$, since the zero solution satisfies $P(t') = 0$. Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$. Therefore, for any solution which does not vanish identically we may write



    $dfrac{dln P}{dt} = dfrac{1}{P}dfrac{dP}{dt} = k cos^2 (rt - Theta); tag 2$



    we integrate 'twixt $t_0$ and $t$:



    $ln dfrac{P(t)}{P(t_0)} = ln P(t) - ln P(t_0) = k displaystyle int_{t_0}^t cos^2(rt - Theta) ; dt; tag 3$



    we have in general (from a table of integrals):



    $displaystyle int cos^2 ax ; dx = dfrac{x}{2} + dfrac{sin 2ax}{4a}; tag 4$



    setting



    $alpha = dfrac{Theta}{r}, tag 5$



    we write



    $rt - Theta = r(t - alpha), tag 6$



    and



    $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = int_{t_0}^t cos^2 r(t - alpha) ; dt = left ( dfrac{t - alpha}{2} + dfrac{sin 2r(t - alpha)}{4r} right vert_{t_0}^t$
    $= dfrac{t - t_0}{2} + dfrac{sin 2r(t - alpha) - sin 2r(t_0 - alpha)}{4r}$
    $= left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r}right ) - left ( dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r} right ); tag 7$



    setting



    $beta(t_0) = dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r}, tag 8$



    we write



    $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} - beta(t_0); tag 9$



    returning to (3),



    $ln dfrac{P(t)}{P(t_0)} = k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0), tag{10}$



    or



    $P(t) = P(t_0) exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0) right )$
    $= P(t_0) e^{ - kbeta(t_0) } exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right )right ); tag{11}$



    there is not much to be gained at this point by furhter re-arrangement of (11). The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.






    share|cite|improve this answer




























      1














      The equation



      $dfrac{dP}{dt} = kP cos^2 (rt - Theta) tag 1$



      is in fact of the variables-seperable type, to wit:



      If $P(t') = 0$ for any $t' in Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t in Bbb R$, since the zero solution satisfies $P(t') = 0$. Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$. Therefore, for any solution which does not vanish identically we may write



      $dfrac{dln P}{dt} = dfrac{1}{P}dfrac{dP}{dt} = k cos^2 (rt - Theta); tag 2$



      we integrate 'twixt $t_0$ and $t$:



      $ln dfrac{P(t)}{P(t_0)} = ln P(t) - ln P(t_0) = k displaystyle int_{t_0}^t cos^2(rt - Theta) ; dt; tag 3$



      we have in general (from a table of integrals):



      $displaystyle int cos^2 ax ; dx = dfrac{x}{2} + dfrac{sin 2ax}{4a}; tag 4$



      setting



      $alpha = dfrac{Theta}{r}, tag 5$



      we write



      $rt - Theta = r(t - alpha), tag 6$



      and



      $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = int_{t_0}^t cos^2 r(t - alpha) ; dt = left ( dfrac{t - alpha}{2} + dfrac{sin 2r(t - alpha)}{4r} right vert_{t_0}^t$
      $= dfrac{t - t_0}{2} + dfrac{sin 2r(t - alpha) - sin 2r(t_0 - alpha)}{4r}$
      $= left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r}right ) - left ( dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r} right ); tag 7$



      setting



      $beta(t_0) = dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r}, tag 8$



      we write



      $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} - beta(t_0); tag 9$



      returning to (3),



      $ln dfrac{P(t)}{P(t_0)} = k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0), tag{10}$



      or



      $P(t) = P(t_0) exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0) right )$
      $= P(t_0) e^{ - kbeta(t_0) } exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right )right ); tag{11}$



      there is not much to be gained at this point by furhter re-arrangement of (11). The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.






      share|cite|improve this answer


























        1












        1








        1






        The equation



        $dfrac{dP}{dt} = kP cos^2 (rt - Theta) tag 1$



        is in fact of the variables-seperable type, to wit:



        If $P(t') = 0$ for any $t' in Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t in Bbb R$, since the zero solution satisfies $P(t') = 0$. Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$. Therefore, for any solution which does not vanish identically we may write



        $dfrac{dln P}{dt} = dfrac{1}{P}dfrac{dP}{dt} = k cos^2 (rt - Theta); tag 2$



        we integrate 'twixt $t_0$ and $t$:



        $ln dfrac{P(t)}{P(t_0)} = ln P(t) - ln P(t_0) = k displaystyle int_{t_0}^t cos^2(rt - Theta) ; dt; tag 3$



        we have in general (from a table of integrals):



        $displaystyle int cos^2 ax ; dx = dfrac{x}{2} + dfrac{sin 2ax}{4a}; tag 4$



        setting



        $alpha = dfrac{Theta}{r}, tag 5$



        we write



        $rt - Theta = r(t - alpha), tag 6$



        and



        $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = int_{t_0}^t cos^2 r(t - alpha) ; dt = left ( dfrac{t - alpha}{2} + dfrac{sin 2r(t - alpha)}{4r} right vert_{t_0}^t$
        $= dfrac{t - t_0}{2} + dfrac{sin 2r(t - alpha) - sin 2r(t_0 - alpha)}{4r}$
        $= left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r}right ) - left ( dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r} right ); tag 7$



        setting



        $beta(t_0) = dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r}, tag 8$



        we write



        $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} - beta(t_0); tag 9$



        returning to (3),



        $ln dfrac{P(t)}{P(t_0)} = k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0), tag{10}$



        or



        $P(t) = P(t_0) exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0) right )$
        $= P(t_0) e^{ - kbeta(t_0) } exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right )right ); tag{11}$



        there is not much to be gained at this point by furhter re-arrangement of (11). The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.






        share|cite|improve this answer














        The equation



        $dfrac{dP}{dt} = kP cos^2 (rt - Theta) tag 1$



        is in fact of the variables-seperable type, to wit:



        If $P(t') = 0$ for any $t' in Bbb R$, then by uniqueness of solutions $P(t) = 0$ for all $t in Bbb R$, since the zero solution satisfies $P(t') = 0$. Furthermore, by if necessary reversing the sign of $P$ (a valid operation by linearity) if necessary we may take $P > 0$. Therefore, for any solution which does not vanish identically we may write



        $dfrac{dln P}{dt} = dfrac{1}{P}dfrac{dP}{dt} = k cos^2 (rt - Theta); tag 2$



        we integrate 'twixt $t_0$ and $t$:



        $ln dfrac{P(t)}{P(t_0)} = ln P(t) - ln P(t_0) = k displaystyle int_{t_0}^t cos^2(rt - Theta) ; dt; tag 3$



        we have in general (from a table of integrals):



        $displaystyle int cos^2 ax ; dx = dfrac{x}{2} + dfrac{sin 2ax}{4a}; tag 4$



        setting



        $alpha = dfrac{Theta}{r}, tag 5$



        we write



        $rt - Theta = r(t - alpha), tag 6$



        and



        $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = int_{t_0}^t cos^2 r(t - alpha) ; dt = left ( dfrac{t - alpha}{2} + dfrac{sin 2r(t - alpha)}{4r} right vert_{t_0}^t$
        $= dfrac{t - t_0}{2} + dfrac{sin 2r(t - alpha) - sin 2r(t_0 - alpha)}{4r}$
        $= left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r}right ) - left ( dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r} right ); tag 7$



        setting



        $beta(t_0) = dfrac{t_0}{2} + dfrac{sin 2r(t_0 - alpha)}{4r}, tag 8$



        we write



        $displaystyle int_{t_0}^t cos^2 (rt - Theta) ; dt = dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} - beta(t_0); tag 9$



        returning to (3),



        $ln dfrac{P(t)}{P(t_0)} = k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0), tag{10}$



        or



        $P(t) = P(t_0) exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right ) - kbeta(t_0) right )$
        $= P(t_0) e^{ - kbeta(t_0) } exp left ( k left ( dfrac{t}{2} + dfrac{sin 2r(t - alpha)}{4r} right )right ); tag{11}$



        there is not much to be gained at this point by furhter re-arrangement of (11). The reader my substitute specific values of the constants (including $t_0$ and $P(t_0)$) to realize a speciic, concrete solution.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '18 at 5:10

























        answered Dec 10 '18 at 4:57









        Robert Lewis

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