# ODE [Pending...]

**Vartek** 17 Jul 2020, 21:35

### 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

**Admin** 19 Jul 2020, 15:11

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

**Vartek** 19 Jul 2020, 15:53

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

**Vartek** 19 Jul 2020, 16:17

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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