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ODE [Pending...]

My question

Hello,

I have a home work in which i have to give the exact solution of a DE and than use it to plot a graph to make a comparison with a numerical solution usiging RK 4th order methode,
my question is how can found the solution of this DE!

thanks for the reply.

Relevant page

1. Solving Differential Equations

What I've done so far

qq e(du)/dt= (1-t)u-u^2 qq

with e=constant (not exponential)

and the initial condition :

qq u(0)=1 qq

X

Hello,

I have a home work in which i have to give the exact solution of a DE and than use it to plot a graph to make a comparison with a numerical solution usiging RK 4th order methode,
my question is how can found the solution of this DE!

thanks for the reply.
Relevant page

<a href="/differential-equations/1-solving-des.php">1. Solving Differential Equations</a>

What I've done so far

qq e(du)/dt= (1-t)u-u^2 qq

with e=constant  (not exponential)

and the initial condition :

qq u(0)=1 qq

Re: ODE

You haven't shown any working so I can't see where you are getting stuck. Your first task is to decide what kind of differential equation it is. Is it separable? An integrable combination? First order linear?

X

You haven't shown any working so I can't see where you are getting stuck. Your first task is to decide what kind of differential equation it is. Is it separable? An integrable combination? First order linear?

Re: ODE

yes sorry i forgot to show where i'am stuck;
so i try to rewrit it as a first order liner and i got it like this and i am not so sure about it :
qq (du)/dt+(1/(1-t))*u=e qq

X

yes sorry i forgot  to show where i'am stuck;
so i try to rewrit it as a first order liner and i got it like this and i am not so sure about it :
qq (du)/dt+(1/(1-t))*u=e qq

Re: ODE

and for now i got to this but i don't understand why after making the integration of p(t) i've go back to qq(1/(1-t))qq

qq p(t)=(1/(1-t)) and Q(t)=e qq
so qq IF= exp(int p(t)dt)= exp(-ln(1-t))=(1/(1-t))qq
qq u*(1/(1-t))=int e (1/(1-t)) dt qq
qq u*(1/(1-t))= -e ln ( 1-t)qq
qq u= -e ln (1-t) *(1-t)qq

this is what i've got so far !

X

and for now i got to this but i don't understand why after making the integration of p(t) i've go back to qq(1/(1-t))qq

qq p(t)=(1/(1-t)) and Q(t)=e qq
so qq IF= exp(int p(t)dt)= exp(-ln(1-t))=(1/(1-t))qq
qq u*(1/(1-t))=int e (1/(1-t)) dt qq
qq u*(1/(1-t))= -e ln ( 1-t)qq
qq u= -e ln (1-t) *(1-t)qq 

this is what i've got so far !

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