Simple integration

[nad_scorp]

How to solve

y ' (t) = (t - y) / 2

[Ambivalentiowa]

You have to seperate the variables first so:

y*t = (t-y)/2

2y*t = t-y

2y+y = t/t

3y = 1

After integrating

3y*dy/dx = t

dy/dx = t/3y

[nad_scorp]

thanx but I didn't really understand how u did it

BTW. y'(t) == derivetive

just 2 check u got it right

[wy125]

nad_scorp:

You can't solve that equation by separation of variables. You need to use an integration factor. These factors are used for equations of the form:

dy/dx = f(x,y)

Rewrite the equation like this

dy/dt +y/2 = t/2

By examining the equation we see the integration factor is given by

exp(Integral(1/2 dt))

which is. exp(0.5 t)

Note: This factor works for cases when f(x,y) is linear in y. If it wasn't you would need a different integration factor.

Anyway, if we multiply the differential equation by our integration factor we get

d/dt[y exp(0.5t)] = 0.5t*exp(0.5t)

We can now solve this directly since the variables are now separated. You'll need to use integration by parts. When you're done you should get:

y(t) = t + (y0 + 2)*exp(-0.5*t) - 2

where y0 is the value of y when t is 0


Hope that helps. You might want to check out a book on ordinary differential equations.

[wy125]

Okay, I thought about the problem some more and it can be solved another way without using an integration factor.

The original equation is

dy/dt = (t - y)/2


If we let

u = t - y


then

du/dt = 1 - dy/dt


using this we get the following new equation


1 - du/dt = u/2

du/dt = 1 - u/2 = (2 - u)/2

This new equation is separable and can be integrated as usual. You'll get the same answer that was found using the integration factor after substituting back in for y. Generally speaking, I prefer the integrating factor, but it's up to you.

[nad_scorp]

THANX MAN


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That's it