Solve this: $frac{dx}{dt} = [a-(b+c)]x-ax^2$ by separation of variables












0












$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31


















0












$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31
















0












0








0





$begingroup$


So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






ordinary-differential-equations







share|cite|improve this question











$endgroup$




So I have this equation



$$dfrac{dx}{dt} = [a-(b+c)]x-ax^2$$



which I need to solve by separation of variables. I've never seen an example like this where you have a sum on the RHS of the eq. Usually the examples are just a product on the RHS.



I have the initial condition of $x(0)=x_0$






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 12 '18 at 12:39









Davide Giraudo

127k16151264




127k16151264










asked Dec 12 '18 at 12:11









countduckulacountduckula

1




1












  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31




















  • $begingroup$
    "Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
    $endgroup$
    – Did
    Dec 12 '18 at 14:46












  • $begingroup$
    True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
    $endgroup$
    – countduckula
    Dec 12 '18 at 15:21










  • $begingroup$
    $$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
    $endgroup$
    – Did
    Dec 13 '18 at 22:31


















$begingroup$
"Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
$endgroup$
– Did
Dec 12 '18 at 14:46






$begingroup$
"Usually the examples are just a product on the RHS." The RHS is a product, namely $xcdot(a-b-c-ax)$.
$endgroup$
– Did
Dec 12 '18 at 14:46














$begingroup$
True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
$endgroup$
– countduckula
Dec 12 '18 at 15:21




$begingroup$
True but I can't totally factor out the x. I'm just being dumb with this tbh. I need to revise D.E. from last term - I've totally forgotten everything lol.
$endgroup$
– countduckula
Dec 12 '18 at 15:21












$begingroup$
$$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
$endgroup$
– Did
Dec 13 '18 at 22:31






$begingroup$
$$frac u{x(u-x)}=frac1x+frac a{u-ax}=left(log|x|-log|u-ax|right)'=left(logleft|frac x{u-ax}right|right)'$$
$endgroup$
– Did
Dec 13 '18 at 22:31












1 Answer
1






active

oldest

votes


















-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49
















-1












$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49














-1












-1








-1





$begingroup$

Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$






share|cite|improve this answer











$endgroup$



Separation of variables means putting everything in $x$ on one side and $t$ on the other, then integrating:



NOTE I will replace $dx$ with $dX$ and $dt$ with $T$ to avoid having $x$ inside and outside the integral sign.



$frac{dX}{dT} = (a-(b+c))X-aX^2 Rightarrow frac{dX}{(a-(b+c))X-aX^2} = dT$
$Rightarrow intlimits_{x0}^xfrac{dX}{(a-(b+c))X-aX^2} =intlimits_0^tdT$
$Rightarrow t=frac{ln(x)-ln((a-(b+c))-ax)}{a-(b+c)}-frac{ln(x_0)-ln((a-(b+c))-ax_0)}{a-(b+c)}$



Finally isolate $x$ to have $x=f(t,a,b,c)$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 12 '18 at 14:43

























answered Dec 12 '18 at 12:27









TheD0ubleTTheD0ubleT

39218




39218












  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49


















  • $begingroup$
    There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
    $endgroup$
    – Satish Ramanathan
    Dec 12 '18 at 12:53










  • $begingroup$
    I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
    $endgroup$
    – TheD0ubleT
    Dec 12 '18 at 14:45










  • $begingroup$
    Why hide carefully the core of the proof?
    $endgroup$
    – Did
    Dec 12 '18 at 14:49
















$begingroup$
There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
$endgroup$
– Satish Ramanathan
Dec 12 '18 at 12:53




$begingroup$
There is some mix up in your results when you do partial fractions. I think you are missing an a in front of the second log term!!
$endgroup$
– Satish Ramanathan
Dec 12 '18 at 12:53












$begingroup$
I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
$endgroup$
– TheD0ubleT
Dec 12 '18 at 14:45




$begingroup$
I did forget the part with $x_0$ but i don't think i'm missing an $a$ (not according to Wolfram Alpha
$endgroup$
– TheD0ubleT
Dec 12 '18 at 14:45












$begingroup$
Why hide carefully the core of the proof?
$endgroup$
– Did
Dec 12 '18 at 14:49




$begingroup$
Why hide carefully the core of the proof?
$endgroup$
– Did
Dec 12 '18 at 14:49


















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