Integrating partial fractions so $x(t)$ is subject

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So I've got $fracdxdt = bx(1-x)$ and I need to find the general solution for $x(t)$.



I've separated it into partial fractions:



$$frac1x(1-x) = frac1x + frac11-x$$



Which leads to $ln(x/1-x) = bt + C$ and then when I apply the exponent function:



$$fracx1-x = e^Ccdot e^bt$$



How can I get $x(t)$ to be the subject? Thanks :)




calculus integration




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    This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
    – Yves Daoust
    Aug 23 at 8:05














up vote
0
down vote

favorite












So I've got $fracdxdt = bx(1-x)$ and I need to find the general solution for $x(t)$.



I've separated it into partial fractions:



$$frac1x(1-x) = frac1x + frac11-x$$



Which leads to $ln(x/1-x) = bt + C$ and then when I apply the exponent function:



$$fracx1-x = e^Ccdot e^bt$$



How can I get $x(t)$ to be the subject? Thanks :)




calculus integration




share|cite|improve this question


















  • 1




    This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
    – Yves Daoust
    Aug 23 at 8:05












up vote
0
down vote

favorite









up vote
0
down vote

favorite











So I've got $fracdxdt = bx(1-x)$ and I need to find the general solution for $x(t)$.



I've separated it into partial fractions:



$$frac1x(1-x) = frac1x + frac11-x$$



Which leads to $ln(x/1-x) = bt + C$ and then when I apply the exponent function:



$$fracx1-x = e^Ccdot e^bt$$



How can I get $x(t)$ to be the subject? Thanks :)




calculus integration




share|cite|improve this question














So I've got $fracdxdt = bx(1-x)$ and I need to find the general solution for $x(t)$.



I've separated it into partial fractions:



$$frac1x(1-x) = frac1x + frac11-x$$



Which leads to $ln(x/1-x) = bt + C$ and then when I apply the exponent function:



$$fracx1-x = e^Ccdot e^bt$$



How can I get $x(t)$ to be the subject? Thanks :)





calculus integration





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edited Aug 23 at 7:51









tarit goswami

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asked Aug 23 at 7:12









Hews

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  • 1




    This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
    – Yves Daoust
    Aug 23 at 8:05












  • 1




    This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
    – Yves Daoust
    Aug 23 at 8:05







1




1




This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
– Yves Daoust
Aug 23 at 8:05




This is elementary algebra, you shouldn't be asking that question ! (I am not trying to be rude, but someone dealing with ODE's should be able.)
– Yves Daoust
Aug 23 at 8:05










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$x(t)=frac e^Ce^bt 1+e^Ce^bt$. To derive this write $x=(1-x)e^Ce^bt=e^Ce^bt-xe^Ce^bt$. Take the last term to the left side to get $x[1+e^Ce^bt]=e^Ce^bt$. Now divide by $1+e^Ce^bt$






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  • Awesome, thank you :)
    – Hews
    Aug 23 at 7:33










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up vote
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accepted










$x(t)=frac e^Ce^bt 1+e^Ce^bt$. To derive this write $x=(1-x)e^Ce^bt=e^Ce^bt-xe^Ce^bt$. Take the last term to the left side to get $x[1+e^Ce^bt]=e^Ce^bt$. Now divide by $1+e^Ce^bt$






share|cite|improve this answer




















  • Awesome, thank you :)
    – Hews
    Aug 23 at 7:33














up vote
1
down vote



accepted










$x(t)=frac e^Ce^bt 1+e^Ce^bt$. To derive this write $x=(1-x)e^Ce^bt=e^Ce^bt-xe^Ce^bt$. Take the last term to the left side to get $x[1+e^Ce^bt]=e^Ce^bt$. Now divide by $1+e^Ce^bt$






share|cite|improve this answer




















  • Awesome, thank you :)
    – Hews
    Aug 23 at 7:33












up vote
1
down vote



accepted







up vote
1
down vote



accepted






$x(t)=frac e^Ce^bt 1+e^Ce^bt$. To derive this write $x=(1-x)e^Ce^bt=e^Ce^bt-xe^Ce^bt$. Take the last term to the left side to get $x[1+e^Ce^bt]=e^Ce^bt$. Now divide by $1+e^Ce^bt$






share|cite|improve this answer












$x(t)=frac e^Ce^bt 1+e^Ce^bt$. To derive this write $x=(1-x)e^Ce^bt=e^Ce^bt-xe^Ce^bt$. Take the last term to the left side to get $x[1+e^Ce^bt]=e^Ce^bt$. Now divide by $1+e^Ce^bt$







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answered Aug 23 at 7:20









Kavi Rama Murthy

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  • Awesome, thank you :)
    – Hews
    Aug 23 at 7:33
















  • Awesome, thank you :)
    – Hews
    Aug 23 at 7:33















Awesome, thank you :)
– Hews
Aug 23 at 7:33




Awesome, thank you :)
– Hews
Aug 23 at 7:33

















 

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