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Separable Differential Equations

Recent questions in Separable Differential Equations

erwachsenc6 2022-10-23

Are both results valid solutions for the separable DE?
I have a separable differential equation (with no initial value)
$d x + \frac{2 z - 1}{2 - z} d z = 0$
And the way my professor did it is
$\int d x = x + C$
$\int \frac{2 z - 1}{2 - z} d z = - \int (\frac{2 (z - 2)}{z - 2} + \frac{3}{z - 2}) d z$
$= - 2 z - 3 l n | z - 2 |$
Resulting in
$x - 2 z - 3 l n | z - 2 | = C_{1}$
The way that he did it is with both integrals on the same side.
What I did was first moved them to separate sides, with the signs changing as necessary.
$d x = \frac{1 - 2 z}{2 - z} d z$
$\frac{1 - 2 z}{2 - z} = \frac{2 (2 - z) - 3}{2 - z}$
So then my integral becomes
$\int \frac{1 - 2 z}{2 - z} d z = 2 \int \frac{2 - z}{2 - z} d z - 3 \int \frac{1}{2 - z} d z$
The result being
$2 z - 3 l n | 2 - z |$
Rearranging to get the final solution I end up with
$x - 2 z + 3 l n | 2 - z | = C_{1}$
The two solutions differ because of the manipulation earlier from changing sides and moving them back afterwards.
Professor's solution:
$x - 2 z - 3 l n | z - 2 | = C_{1}$
My solution:
$x - 2 z + 3 l n | 2 - z | = C_{1}$
My question is, are both of these valid solutions? The domain where ln is defined is different in both solutions. If there is a mistake, what have I done wrong?
Thank you.

Bodonimhk 2022-10-23

How to solve $(1 + x y) y d x + (1 - x y) x d y = 0$ by sepatation, as a linear equation or homogeneous?
We have been taught only these 3 types of problems: separable, linear equations or homogeneous. I tried using these methods, but I don't think this is a homogeneous or a linear differential equation. I tried separating the variables but couldn't. I strongly believe that this can be only solved by separating variables otherwise the question is wrong.

Taniya Melton 2022-10-23

Finding a relationship between x and y of a DE
I have the differential equation:
$\frac{d y}{d x} = \frac{- 5 y - x y}{- 4 x - x y}$
How do I go about finding a relationship between $x$ and $y$ ?

Sonia Elliott 2022-10-22

How to calculate this integral when the function is unknown?
Let $f : [0, 1] \to R$ with $f^{'} (x) = \sqrt{1 + f^{2} (x)}$ for all $x \in [0, 1]$ . If $f (0) + f (1) = 0,$ calculate the integral
$I = \int_{0}^{1} f (x) d x$
Any help would be appreciated. Thanks.

Rubi Garner 2022-10-22

Differential Equation
solve differential equation
$\frac{d y}{d x} = \frac{(3 x - y - 6)}{(x + y + 2)}$
I tried to do this but it´s first order and posible is separable variables

Winston Todd 2022-10-19

Integrating a separable differential equation in differential form
I am studying differential equations from a book called Differential Equations (Schaum's outlines). In it, it says if we have a separable differential equation:
$A (x) d x + B (y) d y = 0, y (x_{0}) = y_{0}$
then, the solution to the initial-value problem can be obtained by
$\int_{x_{0}}^{x} A (x) d x + \int_{y_{0}}^{y} B (y) d y = 0$
And that's my problem. I was expecting something like
$\int_{x_{0}}^{x} A (x) d x + \int_{x_{0}}^{x} B (y) d y = 0$
as we integrate the first equation from $x_{o}$ to $x$
I admit that what i am stating is not so intuitive (in fact, the second equation makes more sense than the third). But how can i prove that
$\int_{x_{0}}^{x} B (y (x)) d y (x) = \int_{y_{0}}^{y} B (y) d y$

Renelyn Delas Alas2022-10-11

Is 4 = {4}?

sexiboi150nc 2022-10-08

How to verify my solution to an separable differential equation?
I have this question:
Find the general solution to the separable differential equation
$\frac{d y}{d x} = y (1 - y) .$

albamarronpe 2022-09-30

Solving a separable differential equation: What's wrong with my calculation?
Solve the following separable differential equation:
$\frac{d y}{d x} = (- 4) \cdot e^{y} \cdot c o s (4 x)$

Joyce Sharp 2022-09-26

Separable differential equation: Particular solutions given initial conditions.
If given initial conditions for a separable differential equation, finding them is easy enough, for example if $y (0) = 0$
But how do I know that these are the only solutions? Or that this solution even exists at all before I start calculating it?

Luke Abiad2022-09-15

A small metal bar, whose initial temperature was 20° C, is dropped into a large container of bowling water. How long will it take the bar to reach 90°C if it is known that its temperature increase to 22°C in 1 second?

Haiphongum 2022-09-15

Show substitution leads to a separable differential equation.
Consider the differential equation...
$y^{'} = f (\frac{y}{t})$
Show that the substitution $v = \frac{v}{t}$ leads to a separable differential equation in v
Here's what I did.
$v = \frac{y}{t}$
$\frac{d v}{d t} = \frac{d y}{d t} - \frac{1}{t^{2}}$
Sub into the orignal.
$\frac{d v}{d t} = f (v) - \frac{1}{t^{2}}$
$\frac{d v}{f (v)} = \frac{1}{t^{2}} d t$
This is where I get stuck. Is this the what the question is asking for, or am I forgetting to do something?

traffig75 2022-09-14

First order separable differential equations
$\frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{x d x} = 0$
Using change of variable equals to $Z = \frac{d y}{d x}$ what is the 1st order separable differential equation for Z as function of x?
and solve for y(x).

Dorsheele0p 2022-08-19

Elementary Separable Differential Equation with tricky integral
Dear Plaimath. The problem is to solve the separable differential equation $\sqrt{x y} \frac{d y}{d x} = \sqrt{4 - x}$
My work thus far: $\sqrt{x y} d y = \sqrt{4 - x} d x \sqrt{y} d y = \sqrt{\frac{4 - x}{x}} d x \int \sqrt{y} d y = \int \sqrt{\frac{4 - x}{x}} d x \frac{2}{3} y^{3 / 2} = \int \sqrt{\frac{4 - x}{x}} d x \frac{2}{3} y^{3 / 2} = \int \sqrt{\frac{4}{x} - 1} d x$
Is there an easier approach I have missed, or should I persist in trying to solve the integral.

Makayla Eaton 2022-08-17

Solve the following differential equation using separation of variables
$f^{'} (x) + 2 x [f^{2} (x) - f (x)] = 0$

rivasguss9 2022-08-16

Solving a separable differential equations
We have an equation:
$\frac{d P}{d t} = k P (1 - \frac{P}{M}),$
and we need to find $P (t)$ given the initial $P (0)$ given the initial $P (0)$ . Here k is a constant and P represents population, M represents maximum population.
I tried to use separable differential equations, but I am slightly confused. There is a hint:
$\frac{M}{P (M - P)} = \frac{1}{P} + \frac{1}{M - P},$
and I need to simplify as far as possible. I have tried to use separable equations but I cannot seem to get the hint equation. I think the main issue is trying to get all P's on to one side.
Thank you in advance!

dredyue 2022-08-12

Why does the method of separable equations work in differential equations?
In differential equations, the idea of multiplying by an infinitesimal, dx, is used in the method of separable equations. My confusion is in why this works as I've heard many times in the past that you shouldn't multiply and cancel out infinitesimals like that. I understand that in the setting of nonstandard analysis that there may not be something wrong with this, but in the usual setting is there a more rigorous interpretation of this?

Dean Summers 2022-07-25

Consider the equation $d y / d x = y^{1 / 3}$ .

Deborah Wyatt 2022-07-23

Having trouble solving this separable differential equation
I am having some trouble with the following separable differential equation
$\frac{d x}{d t} = x (x - 1) (x - 3)$
with initial condition $x (0) = 2$ . What is $lim_{t \to \infty} x (t)$ ?
I am having some trouble with the logarithmic laws when solving for $x (t)$

Hayley Bernard 2022-07-23

Solving separable differential equations by formal computations
When solving a separable differential equation we move from the original equation
$N (y) \frac{d y}{d x} = M (x)$
to
$N (y) d y = M (x) d x$
Then we integrate both sides. My question is how precise is the expression $N (y) d y = M (x) d x$ ? is it a formal writing to simplify computation and get quickly into integrating both sides or is it a precise mathematical expression that has a precise mathematical meaning but goes beyond an introductory course on differential equations? Thanks for your help !

Separable differential equations are a type of first-order ordinary differential equation (ODE) that can easily be solved by rearranging terms. The goal is to move all terms with the independent variable to one side of the equation and all terms with the dependent variable to the other side. This allows for the equation to be separated into two parts and solved for the unknown variable. In addition to being easier to solve than other first-order ODEs, separable equations can be used to model many real-world phenomena. This can include things such as the rate of population growth, the flow of heat, and the motion of a spring. With the right tools, these equations can be used to gain valuable insights into a variety of problems.

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