Help with DE problem and seperation of variables

[JasonJo]

i found the DE for a problem, and it was y'(t) = .06 - y/1040

and then the problem gave me the following hint:

Find the amount of sugar after t minutes. Note: When you solve by separation of variables, keep the coefficient of y on the right side and bring over to the denominator on the left side an expression of the form (y - some constant).

what does this mean? can anyone actually show me how to do this?? thanks

 

[AKG]

EDIT: Of course, you can do it with separation of variables, I just wasn't thinking. See StatusX's post below.

1) Take the derivative again, so y'' = (-1/1040)y'. You get y' = Ce(-t/1040), and you can integrate this to find y = De(-t/1040) + E, and then plug this back into the original DE and solve for D in terms of E, or vice versa. You will be left with one unknown, since no initial conditions are specified.

2) Find the solution to the associated homogeneous equation, y'(t) = -y/1040, let's call that solution ch(t), where c is a co-efficient that can vary (i.e. ch(t) will be a solution to the homogeneous equation for any real c). Then, find any particular solution p(t). The constant function p(t) = 1040/0.06 will do the trick. Your answer will be y = ch + p.

This is generally true. If you have an n-th order differential equation (so it includes the n-th derivative, whereas this is a 1st order equation since it only includes up to the 1st derivative), you will have n linearly independent homogeneous solutions, h1, h2, ..., hn. Then, if you know any particular solution p, the general solution will be:

c1h1 + c2h2 + ... + cnhn + p

for any choice of c1, ..., cn. This is obviously a family of solutions, but if you look at all possible choices of c1, ..., cn, then this will give you all possible solutions.

3) Look at the equation as y'(t) + a(t)y(t) = b(t), where a(t) = 1/1040 and b(t) = 0.6. Your solution will be:

\frac{\int u(x)b(x)\, dx + C}{u(x)}*

where

u(x) = \exp \left (\int a(x)\, dx\right )

This is also true in general, if you have an equation of the form y' + ay = b. All solutions will be defined by * for all choices of C.

 

[StatusX]

dy/dt = 0.06 - y/1040

\frac{dy}{0.06 - y/1040} = dt

\int \frac{dy}{0.06 - y/1040} = \int dt

-1040 ln(0.06 - y/1040) = t + C

and so on