First order differential equations with 2 varriables












0












$begingroup$


Does anybody know how to solve the following differntial equation?



$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$



where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant



EDIT:

This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.



Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.



The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$



From circular motion we know that $a_r = omega ^2 r$

where $r$ is the radius, and $omega$



Further we know that $v=omega*r$, therefore:
$a_s=frac{d}{dt}(omega r)$



For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$



Further usefull information:

Length of the track covered at a certain time
$l=v*t=omega r t$



With all this the radius will be a function of time, $r(t)$

and the angular velocity will be a function of time, $omega (t)$



Boundary conditions:
$r(0) = 0$
$omega (0) = 0$



Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$



Here is a diagram for clarity:
Constant acceleration spiral track






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    One more differential equation is needed to solve it.
    $endgroup$
    – Narasimham
    Dec 29 '18 at 20:03
















0












$begingroup$


Does anybody know how to solve the following differntial equation?



$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$



where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant



EDIT:

This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.



Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.



The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$



From circular motion we know that $a_r = omega ^2 r$

where $r$ is the radius, and $omega$



Further we know that $v=omega*r$, therefore:
$a_s=frac{d}{dt}(omega r)$



For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$



Further usefull information:

Length of the track covered at a certain time
$l=v*t=omega r t$



With all this the radius will be a function of time, $r(t)$

and the angular velocity will be a function of time, $omega (t)$



Boundary conditions:
$r(0) = 0$
$omega (0) = 0$



Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$



Here is a diagram for clarity:
Constant acceleration spiral track






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    One more differential equation is needed to solve it.
    $endgroup$
    – Narasimham
    Dec 29 '18 at 20:03














0












0








0





$begingroup$


Does anybody know how to solve the following differntial equation?



$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$



where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant



EDIT:

This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.



Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.



The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$



From circular motion we know that $a_r = omega ^2 r$

where $r$ is the radius, and $omega$



Further we know that $v=omega*r$, therefore:
$a_s=frac{d}{dt}(omega r)$



For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$



Further usefull information:

Length of the track covered at a certain time
$l=v*t=omega r t$



With all this the radius will be a function of time, $r(t)$

and the angular velocity will be a function of time, $omega (t)$



Boundary conditions:
$r(0) = 0$
$omega (0) = 0$



Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$



Here is a diagram for clarity:
Constant acceleration spiral track






ordinary-differential-equations







share|cite|improve this question











$endgroup$




Does anybody know how to solve the following differntial equation?



$$w^2cdot r+frac{dw}{dt}cdot r+frac{dr}{dt}cdot w=3g$$



where w and r are a function of time so $w(t)$ and $r(t)$.
$g$ is a constant



EDIT:

This equation came from a thought problem of mine which I will now explain since I still do not know how to approach the problem.



Basically I want to design a spiral track in which a vehicle will accelerate. However the total acceleration needs to be constant.



The vehicle will have 2 acceleration components, a radial component ($a_r$) and a change in speed component ($a_s$). Since we want constant acceleration:
$a_r + a_s = constant$



From circular motion we know that $a_r = omega ^2 r$

where $r$ is the radius, and $omega$



Further we know that $v=omega*r$, therefore:
$a_s=frac{d}{dt}(omega r)$



For constant acceleration $a_{total} = a_r + a_s = omega ^2 r + frac{d}{dt}(omega r) =constant$



Further usefull information:

Length of the track covered at a certain time
$l=v*t=omega r t$



With all this the radius will be a function of time, $r(t)$

and the angular velocity will be a function of time, $omega (t)$



Boundary conditions:
$r(0) = 0$
$omega (0) = 0$



Now I would like to find a possible set of solutions to create a track by implementing end boundary conditions that i choose e.g. $v(t_{end})=100 [m/s]$ and $l(t_{end})=5000 [m]$



Here is a diagram for clarity:
Constant acceleration spiral track






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 30 '18 at 10:51







Alessandro Simonelli

















asked Dec 29 '18 at 12:02









Alessandro SimonelliAlessandro Simonelli

113




113












  • $begingroup$
    One more differential equation is needed to solve it.
    $endgroup$
    – Narasimham
    Dec 29 '18 at 20:03


















  • $begingroup$
    One more differential equation is needed to solve it.
    $endgroup$
    – Narasimham
    Dec 29 '18 at 20:03
















$begingroup$
One more differential equation is needed to solve it.
$endgroup$
– Narasimham
Dec 29 '18 at 20:03




$begingroup$
One more differential equation is needed to solve it.
$endgroup$
– Narasimham
Dec 29 '18 at 20:03










2 Answers
2






active

oldest

votes


















0












$begingroup$

Calling $z(t) = w(t) r(t)$ we have



$$
z^2(t)/r(t)+z'(t) =3g
$$



now knowing $r(t)$ you can solve for $z(t)$ recovering $w(t) = z(t)/r(t)$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
    $endgroup$
    – Alessandro Simonelli
    Dec 29 '18 at 14:44










  • $begingroup$
    A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
    $endgroup$
    – Cesareo
    Dec 29 '18 at 16:16



















1












$begingroup$

Suppose that $w$ is an arbitrary differentiable function. For any such function your equation is a linear differential equation for $r$. Solve it, say, by the variations of constants formula.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Calling $z(t) = w(t) r(t)$ we have



    $$
    z^2(t)/r(t)+z'(t) =3g
    $$



    now knowing $r(t)$ you can solve for $z(t)$ recovering $w(t) = z(t)/r(t)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
      $endgroup$
      – Alessandro Simonelli
      Dec 29 '18 at 14:44










    • $begingroup$
      A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
      $endgroup$
      – Cesareo
      Dec 29 '18 at 16:16
















    0












    $begingroup$

    Calling $z(t) = w(t) r(t)$ we have



    $$
    z^2(t)/r(t)+z'(t) =3g
    $$



    now knowing $r(t)$ you can solve for $z(t)$ recovering $w(t) = z(t)/r(t)$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
      $endgroup$
      – Alessandro Simonelli
      Dec 29 '18 at 14:44










    • $begingroup$
      A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
      $endgroup$
      – Cesareo
      Dec 29 '18 at 16:16














    0












    0








    0





    $begingroup$

    Calling $z(t) = w(t) r(t)$ we have



    $$
    z^2(t)/r(t)+z'(t) =3g
    $$



    now knowing $r(t)$ you can solve for $z(t)$ recovering $w(t) = z(t)/r(t)$






    share|cite|improve this answer









    $endgroup$



    Calling $z(t) = w(t) r(t)$ we have



    $$
    z^2(t)/r(t)+z'(t) =3g
    $$



    now knowing $r(t)$ you can solve for $z(t)$ recovering $w(t) = z(t)/r(t)$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 29 '18 at 13:56









    CesareoCesareo

    8,9933516




    8,9933516












    • $begingroup$
      Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
      $endgroup$
      – Alessandro Simonelli
      Dec 29 '18 at 14:44










    • $begingroup$
      A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
      $endgroup$
      – Cesareo
      Dec 29 '18 at 16:16


















    • $begingroup$
      Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
      $endgroup$
      – Alessandro Simonelli
      Dec 29 '18 at 14:44










    • $begingroup$
      A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
      $endgroup$
      – Cesareo
      Dec 29 '18 at 16:16
















    $begingroup$
    Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
    $endgroup$
    – Alessandro Simonelli
    Dec 29 '18 at 14:44




    $begingroup$
    Thanks for the answer, but what if I want to find the combination of w(t) and r(t) for which the equation is true?
    $endgroup$
    – Alessandro Simonelli
    Dec 29 '18 at 14:44












    $begingroup$
    A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
    $endgroup$
    – Cesareo
    Dec 29 '18 at 16:16




    $begingroup$
    A differential equation is an equation: so think what do you can ask to an equation like $w(t)+r(t) = 3g$ regarding the values for $w(t)$ or $r(t)$ considered as unknowns. This equation (one) with two unknowns would be undecidable.
    $endgroup$
    – Cesareo
    Dec 29 '18 at 16:16











    1












    $begingroup$

    Suppose that $w$ is an arbitrary differentiable function. For any such function your equation is a linear differential equation for $r$. Solve it, say, by the variations of constants formula.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Suppose that $w$ is an arbitrary differentiable function. For any such function your equation is a linear differential equation for $r$. Solve it, say, by the variations of constants formula.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose that $w$ is an arbitrary differentiable function. For any such function your equation is a linear differential equation for $r$. Solve it, say, by the variations of constants formula.






        share|cite|improve this answer









        $endgroup$



        Suppose that $w$ is an arbitrary differentiable function. For any such function your equation is a linear differential equation for $r$. Solve it, say, by the variations of constants formula.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 29 '18 at 12:42









        Gerhard S.Gerhard S.

        1,06529




        1,06529






























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