Help with DE problem and seperation of variables

In summary, the problem involves finding the amount of sugar after a certain amount of time, using the given differential equation and hint. The solution can be found using separation of variables, by finding the solution to the associated homogeneous equation and a particular solution, or by using the general solution formula for equations of the form y' + ay = b.
  • #1
JasonJo
429
2
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
 
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  • #2
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:

[tex]\frac{\int u(x)b(x)\, dx + C}{u(x)}[/tex]*

where

[tex]u(x) = \exp \left (\int a(x)\, dx\right )[/tex]

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.
 
Last edited:
  • #3
[tex]dy/dt = 0.06 - y/1040 [/tex]

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

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

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

and so on
 

Related to Help with DE problem and seperation of variables

1. What is a DE problem?

A DE (differential equation) problem is a mathematical equation that involves an unknown function and its derivatives. It is used to model various phenomena in science and engineering.

2. How can I solve a DE problem?

One common method for solving DE problems is through separation of variables. This involves isolating the variables on either side of the equation and then integrating both sides to find the solution.

3. What is separation of variables?

Separation of variables is a technique used to solve DE problems by isolating the variables on either side of the equation and then integrating both sides to find the solution.

4. When is separation of variables applicable?

Separation of variables is applicable when the DE problem is a first-order, separable equation. This means that the equation can be written as two separate functions multiplied together.

5. Are there any limitations to using separation of variables?

Yes, there are some limitations to using separation of variables. It can only be used for first-order, separable equations and it may not always provide the most accurate solution. Other methods, such as numerical approximation, may be needed for more complex DE problems.

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