How to solve ode 1st order question
- Thread starter PapayaC
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Hello, and welcome to FMH! 
It is a separable ODE, but you don't have to treat it as such. However, let's go that route, and write:
[MATH]a\d{y}{x}=b-y[/MATH]
Since we are told \(y\ne b\) we may divide through by \(b-y\) without losing a trivial solution. What do you get when doing that?
It is a separable ODE, but you don't have to treat it as such. However, let's go that route, and write:
[MATH]a\d{y}{x}=b-y[/MATH]
Since we are told \(y\ne b\) we may divide through by \(b-y\) without losing a trivial solution. What do you get when doing that?
Ohhh okay so:Hello, and welcome to FMH!
It is a separable ODE, but you don't have to treat it as such. However, let's go that route, and write:
[MATH]a\d{y}{x}=b-y[/MATH]
Since we are told \(y\ne b\) we may divide through by \(b-y\) without losing a trivial solution. What do you get when doing that?
(a/(b-y))dy= dx
a ln(b-y)= x
ln(b-y) = x/a
How do I subject y with the help of e and get rid of b at the same time?
(Also thank you so much for the help)
As Mark explained, given that a is a non-zero constant and b is a constant not in the range of y, we can determine that we have a separable differential equation and solve appropriately. In terms of looking for separable equations, I like Leibniz notation. But you made two mistakes unless you know that y < b. Absolute value and these are indefinite integrals so there are constants.
[MATH]a * \dfrac{dy}{dt} + y = b \implies a * \dfrac{dy}{dt} = b - y \implies \\ a \ dy = (b - y) \ dt \implies a * \dfrac{1}{b - y} \ dy = dt \implies \\ a * \int \ \dfrac{1}{b - y} \ dy = \int dt \implies a * ln|b - y| + c_1 = t + c_2.[/MATH]Now there are quite a few games to play with those constants before taking advantage of latest Mark's hint.
You just said that y is not equal to b. If that is all you know, it might be easier to deal with y > b and y < b as seperate cases.
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@PapayaC
Something occurred to me. You may not have given us the entire problem. We ask that you give us the entire problem in Read Before Posting, but many do not bother to read that. Not giving the entire problem almost always ends up in delaying our giving useful help.
If you were given, as well you may have, additional information, we may be able to deal with the absolute value complication, the constants issue, or both before they make things too complicated.
In general (always???), the solution to a differential equation is not a function, but a family of functions, and we need additional information to determine which of the family is the answer wanted.
In the future, please provide a complete and exact statement of the problem. (If you did so this time, congratulations, but you will not get a specific function as an answer if that is the way the problem was posed.)
Something occurred to me. You may not have given us the entire problem. We ask that you give us the entire problem in Read Before Posting, but many do not bother to read that. Not giving the entire problem almost always ends up in delaying our giving useful help.
If you were given, as well you may have, additional information, we may be able to deal with the absolute value complication, the constants issue, or both before they make things too complicated.
In general (always???), the solution to a differential equation is not a function, but a family of functions, and we need additional information to determine which of the family is the answer wanted.
In the future, please provide a complete and exact statement of the problem. (If you did so this time, congratulations, but you will not get a specific function as an answer if that is the way the problem was posed.)
Oh my bad, it was to find the family not a particular solution so yeah but I'll keep what you said in mind