ODE - Ordinary Differential Equations (1 Viewer)

[Forbidden.]

How does one solve dy/dt = α ln(K / y) y given that y = y₀ and t = 0

This topic is used for many practical applications but this topic annoys me when I can't set up one for a given problem ...

e.g

It is estimated that the population growth rate of a certain developing country will
fall linearly from 2% per year to 1.5% per year over the next decade.

a) Express the population growth rate as a function of time.

The differential equation becomes dy/dt = (0.02 − 0.0005t) y

Rock On

 

[lolokay]

can't you just separate variables + integrate through?

for the question, you know that at t=0, dy/dt = 2%y = 0.02y
and the rate falls at a rate of (2-1.5)%t/10 = 0.05%t = 0.00005t
which gives you the given equation, and if you want to solve for population as a function of time just separate variables:

dy/y = (0.02 - 0.0005t)dt
ln y = 0.02t - 0.00025 t² + C
y = Ke^{0.02t - 0.00025 t2}
and at t=0, y = y0
so y = y0 e^{0.02t - 0.00025 t2}

I'm not sure if I've answered what you're asking though

 

[Forbidden.]

I 1/y dy = I 0.02 − 0.0005t dt

ln(y) = 0.02t - 0.0005t^2/2 + C

t = 0, y = y₀

ln(y) = 0.02t - 0.0005t^2/2 + ln(y₀)

y = exp[0.02t - 0.0005t^2/2 + ln(y₀)]

I could solve this ODE once I knew it but it's generating the ODE from interpreting the question which is the real problem as the next poster's reply was the one I was looking for.

can't you just separate variables + integrate through?

for the question, you know that at t=0, dy/dt = 2%y = 0.02y
and the rate falls at a rate of (2-1.5)%t/10 = 0.05%t = 0.00005t
which gives you the given equation, and if you want to solve for population as a function of time just separate variables:

dy/y = (0.02 - 0.0005t)dt
ln y = 0.02t - 0.00025 t² + C
y = Ke^{0.02t - 0.00025 t2}
and at t=0, y = y0
so y = y0 e^{0.02t - 0.00025 t2}

I'm not sure if I've answered what you're asking though

Thanks, this was it.

But now Second-Order non-homogeneous ODEs are now the next thing driving me up the wall, e.g.

y′′ + 2y′ + 2y = 10 cos 2x
y′ + 3y = 10e^2x

 

[lolokay]

for y′ + 3y = 10e2x:

y'e^3x + 3e^3x y = 10e^5x (multiplying through by integrating factor)
ye^3x = 2e^5x + C (integrating each side)
y = 2e^2x + Ce^-3x

 

[Iruka]

You can solve that DE using Laplace transforms (if you know about Laplace transforms).

Otherwise, you have to solve the homogeneous equation using the characterisitic equation. (I assume that they've taught you how to do that at uni.)

Then you have to find a particular solution.

let y_p be the particular solution.

then y_p = Asin2x + b cos2x for some constants A and B.
Find y'_p and y''_p and substitute back into the non-homogeneous DE. You should be able to work out two simultaneous equations that will tell you what A and B are. (It is sort of like equating coefficients with polynomials.)

The final solution is the sum of both the solutions (the solution to the homogeneous equation and the particular solution.)

 

[shaon0]

Everyone says DEs are very easy to solve. Is this true?

 

[lolokay]

yeah the solvable ones seem pretty easy
apparently there's only really a few ways to solve them, and the rest is putting them in the solvable forms
I think a large part of it though is what to do when you come across ones that can't be solved (much more common I think)

anywayz, you should learn them

 

[shaon0]

yeah the solvable ones seem pretty easy
apparently there's only really a few ways to solve them, and the rest is putting them in the solvable forms
I think a large part of it though is what to do when you come across ones that can't be solved (much more common I think)

anywayz, you should learn them

Where can i learn them from? Any links?

 

[lolokay]

http://www.intmath.com/Differential-equations/DEs-intro.php - for an introduction

http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring-2006/CourseHome/ - free MIT lecture videos + notes

 

[shaon0]

uni

I think they are putting DEs into the 2009 MX2 syllabus.

 

[shaon0]

http://www.intmath.com/Differential-equations/DEs-intro.php - for an introduction

http://ocw.mit.edu/OcwWeb/Mathematics/18-03Spring-2006/CourseHome/ - free MIT lecture videos + notes

Thanks for the links, lolokay

 

[Iruka]

That is one of the suggestions for the new syllabus.

Linear ODEs are not too bad. (Meaning that there is a well developed theory for solving them.)

There are also lots of methods for solving them numerically in the cases when you can't get an explicit solution.

 

[shaon0]

That is one of the suggestions for the new syllabus.

Linear ODEs are not too bad. (Meaning that there is a well developed theory for solving them.)

There are also lots of methods for solving them numerically in the cases when you can't get an explicit solution.

I think they should put DEs in.

 

[Forbidden.]

for y′ + 3y = 10e2x:

y'e^3x + 3e^3x y = 10e^5x (multiplying through by integrating factor)
ye^3x = 2e^5x + C (integrating each side)
y = 2e^2x + Ce^-3x

I think I forgot to put the second-order derivative notation ... because that question belonged to the 2nd-order ODE section.
Otherwise I see whether it's solvable as a Separable, Linear or Exact ODE.

You can solve that DE using Laplace transforms (if you know about Laplace transforms).

Otherwise, you have to solve the homogeneous equation using the characterisitic equation. (I assume that they've taught you how to do that at uni.)

Then you have to find a particular solution.

let y_p be the particular solution.

then y_p = Asin2x + b cos2x for some constants A and B.
Find y'_p and y''_p and substitute back into the non-homogeneous DE. You should be able to work out two simultaneous equations that will tell you what A and B are. (It is sort of like equating coefficients with polynomials.)

The final solution is the sum of both the solutions (the solution to the homogeneous equation and the particular solution.)

Provided I don't fail maths this semester I can do engineering maths next year which has Laplace transforms in it.
Thanks for the y_P method though.

I think they are putting DEs into the 2009 MX2 syllabus.

About time?
Partial differentiation shouldn't be too hard either.

 

[Iruka]

Yeah, Laplace transforms are good fun. They turn a problem involving calculus into one that only involves algebra.

The strange thing at UNSW is that you can do a whole maths degree without learning any transform methods. I think they only teach them in PDEs and some of the engineering maths subjects (which you have to be an engineering student to enroll in.)