Having trouble understanding a question (differential equations)

, I'm not too sure what to do for the below question.

Do I put x=3 and y=0 for the first equation, then x=18 and y=0? I can't seem to understand on how to start as I can't find any examples.

(edited 10 years ago)

Reply 1

ztibor

Original post by kerrypop

, I'm not too sure what to do for the below question.

Do I put x=3 and y=0 for the first equation, then x=18 and y=0? I can't seem to understand on how to start as I can't find any examples.

Put y=0 and arrange the equation

\displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

Integrate and find x=f(t)

Reply 2

kerrypop

Original post by ztibor

Put y=0 and arrange the equation

\displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

Integrate and find x=f(t)

Thanks for the reply :smile:

I still don't really understand what is going on. If I integrate it I should get my answer, what does that mean in terms of the question though?

Reply 3

ztibor

Original post by kerrypop

Thanks for the reply :smile:

I still don't really understand what is going on. If I integrate it I should get my answer, what does that mean in terms of the question though?

For the LHS

With partial fractions

\displaystyle \frac{A}{x}+\frac{B}{12-x}=\frac{12A-Ax+Bx}{x(12-x)}=\frac{1}{x(12-x)}

So
A=B and 12A=1 -> A=B=1/12

\frac{1}{12} \int \left ( \frac{1}{x}+\frac{1}{12-x}\right ) dx=\frac{1}{12}\int dt

\ln \left |\frac{x}{12-x}\right |=t+C

\frac{x}{12-x}=c\cdot e^{t}

You will get the value of c from the initial conditions t=0 -> x_0=3 or 18

Reply 4

kerrypop

Original post by ztibor

For the LHS

With partial fractions

\displaystyle \frac{A}{x}+\frac{B}{12-x}=\frac{12A-Ax+Bx}{x(12-x)}=\frac{1}{x(12-x)}

So
A=B and 12A=1 -> A=B=1/12

\frac{1}{12} \int \left ( \frac{1}{x}+\frac{1}{12-x}\right ) dx=\frac{1}{12}\int dt

\ln \left |\frac{x}{12-x}\right |=t+C

\frac{x}{12-x}=c\cdot e^{t}

You will get the value of c from the initial conditions t=0 -> x_0=3 or 18

Thanks, that has been really helpful :smile:

I can see by using

\frac{dx}{dt}= x\left(1- \frac{x}{12}\right)

I can get to the same answer as you do.

Could you please explain how you started with

\displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

as my algebra isn't that great and I can't see how you got it.

(edited 10 years ago)

Reply 5

brianeverit

Original post by kerrypop

Thanks, that has been really helpful :smile:

I can see by using

\frac{dx}{dt}= x\left(1- \frac{x}{12}\right)

I can get to the same answer as you do.

Could you please explain how you started with

\displaystyle \frac{1}{x\cdot (12-x)} dx=\frac{1}{12} dt

as my algebra isn't that great and I can't see how you got it.

Separation of variables.

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