Recent questions in Separable Differential Equations

Differential EquationsAnswered question

nepojamanuszc 2023-03-23

1 degree on celsius scale is equal to

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scale

A) $\frac{9}{5}$ degree on fahrenheit scale

B) $\frac{5}{9}$ degree on fahrenheit scale

C) 1 degree on fahrenheit scale

D) 5 degree on fahrenheit scale

Differential EquationsAnswered question

goorst9Bi 2022-11-23

Correct approach to this homogeneous differential equation

I am trying to find the general solution to the following equation, but the integral at the end is very complicated and leads me to believe I may have made a mistake somewhere.

$x{y}^{2}\frac{dy}{dx}={y}^{3}+x{y}^{2}-{x}^{2}y-{x}^{3}$

Which, by the substitution $z=\frac{y}{x}$, can be rearranged into the equation

$x\frac{dz}{dx}=1-\frac{1}{z}-\frac{1}{{z}^{2}}$

This is a separable equation, which I separated into

$\int \frac{1}{1-\frac{1}{z}-\frac{1}{{z}^{2}}}dz=\int \frac{1}{x}dx$

The right-hand side is easy to solve, but the left-hand integral is giving me trouble. Assuming I did the steps leading up to it correctly, the integral has me stumped. Even WolframaAlpha is unhelpful. My first thought would be to try a partial fraction, but after a few attempts it does not seem to work.

Am I approaching this differential equation correctly? Is there an error I haven't caught?

I am trying to find the general solution to the following equation, but the integral at the end is very complicated and leads me to believe I may have made a mistake somewhere.

$x{y}^{2}\frac{dy}{dx}={y}^{3}+x{y}^{2}-{x}^{2}y-{x}^{3}$

Which, by the substitution $z=\frac{y}{x}$, can be rearranged into the equation

$x\frac{dz}{dx}=1-\frac{1}{z}-\frac{1}{{z}^{2}}$

This is a separable equation, which I separated into

$\int \frac{1}{1-\frac{1}{z}-\frac{1}{{z}^{2}}}dz=\int \frac{1}{x}dx$

The right-hand side is easy to solve, but the left-hand integral is giving me trouble. Assuming I did the steps leading up to it correctly, the integral has me stumped. Even WolframaAlpha is unhelpful. My first thought would be to try a partial fraction, but after a few attempts it does not seem to work.

Am I approaching this differential equation correctly? Is there an error I haven't caught?

Differential EquationsAnswered question

siotaody 2022-11-23

Separation of variables differential equation

Solve this equation by separation of variables: xu'=3u.

I see that the answer should be u(x)=$c{x}^{3}$, but since I have no knowledge of differential equations, can someone provide the steps?

Solve this equation by separation of variables: xu'=3u.

I see that the answer should be u(x)=$c{x}^{3}$, but since I have no knowledge of differential equations, can someone provide the steps?

Differential EquationsAnswered question

Juan Lowe 2022-11-23

How to solve such fraction differential equation?

Here's my first-order differential equation:

$({x}^{3}-2x{y}^{2})dx+3y{x}^{2}dy=xdy-ydx$

I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

Here's my first-order differential equation:

$({x}^{3}-2x{y}^{2})dx+3y{x}^{2}dy=xdy-ydx$

I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

Differential EquationsAnswered question

quakbIi 2022-11-22

How can i solve this separable differential equation?

Given Problem is to solve this separable differential equation:

${y}^{\mathrm{\prime}}=\frac{y}{4x-{x}^{2}}.$

My approach: was to build the integral of y':

$\int {y}^{\mathrm{\prime}}=\int \frac{y}{4x-{x}^{2}}dy=\frac{{y}^{2}}{2(4x-{x}^{2})}.$

But now i am stuck in differential equations, what whould be the next step? And what would the solution looks like? Or is this already the solution? I doubt that.

P.S. edits were only made to improve language and latex

Given Problem is to solve this separable differential equation:

${y}^{\mathrm{\prime}}=\frac{y}{4x-{x}^{2}}.$

My approach: was to build the integral of y':

$\int {y}^{\mathrm{\prime}}=\int \frac{y}{4x-{x}^{2}}dy=\frac{{y}^{2}}{2(4x-{x}^{2})}.$

But now i am stuck in differential equations, what whould be the next step? And what would the solution looks like? Or is this already the solution? I doubt that.

P.S. edits were only made to improve language and latex

Differential EquationsAnswered question

Kayley Dickson 2022-11-21

Separable Differential Equation

How would you solve the next ODE?

$\frac{dy}{dt}=\frac{at+by+m}{ct+dy+n},$

where $a,b,c,d,m,n$ are constants and $ad=bc$

Corrected.

How would you solve the next ODE?

$\frac{dy}{dt}=\frac{at+by+m}{ct+dy+n},$

where $a,b,c,d,m,n$ are constants and $ad=bc$

Corrected.

Differential EquationsAnswered question

Nicholas Hunter 2022-11-21

Is $(2x+y)dx-xdy=0$ a separable differential equation?

I was given the following differential equation in an assignment the other day:

$(2x+y)dx-xdy=0$

The problem specified to solve the equation using the method of separation of variables. My problem was setting the integral, I tried multiple manipulations with but nothing seemed to work. So, I have to ask can this equation be solved using separation of variables?

I was given the following differential equation in an assignment the other day:

$(2x+y)dx-xdy=0$

The problem specified to solve the equation using the method of separation of variables. My problem was setting the integral, I tried multiple manipulations with but nothing seemed to work. So, I have to ask can this equation be solved using separation of variables?

Differential EquationsAnswered question

szklanovqq 2022-11-19

Completing partial derivatives to make them converge

For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows:

$\frac{\mathrm{\partial}f(x,y)}{\mathrm{\partial}x}=\frac{1}{g(x,y)\sqrt{1-(\frac{ky}{{x}^{(1/3)}}{)}^{2}}}$

$\frac{\mathrm{\partial}f(x,y)}{\mathrm{\partial}y}=\left(\frac{3x}{4}\right){\left(\frac{k}{{x}^{(1/3)}}\right)}^{2}(2y)\frac{1}{g(x,y)\sqrt{1-(\frac{ky}{{x}^{(1/3)}}{)}^{2}}}$

Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0)=x$

I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$. Needless to say, I am not asking that they verify those suggestions, but in case someone would like, these are the inputs to Wolfram integrator:

$\frac{1}{(g(x,y\text{as}r)}\sqrt{1-(kr/{x}^{(}1/3){)}^{2})}\phantom{\rule{0ex}{0ex}}\frac{(3t/4)(k/{t}^{1/3}{)}^{2}(2x)}{(g(x\text{as}t,y\text{as}x)\sqrt{1-(kx/{t}^{(}1/3){)}^{2})}}$

For a function $f(x,y)$ of two independent variables we have an incomplete specification of its partial derivatives as follows:

$\frac{\mathrm{\partial}f(x,y)}{\mathrm{\partial}x}=\frac{1}{g(x,y)\sqrt{1-(\frac{ky}{{x}^{(1/3)}}{)}^{2}}}$

$\frac{\mathrm{\partial}f(x,y)}{\mathrm{\partial}y}=\left(\frac{3x}{4}\right){\left(\frac{k}{{x}^{(1/3)}}\right)}^{2}(2y)\frac{1}{g(x,y)\sqrt{1-(\frac{ky}{{x}^{(1/3)}}{)}^{2}}}$

Problem: finding a suitable $g(x,y)$ that makes the partial derivatives converge to a single function $f(x,y)$ that fulfills the condition $f(x,0)=x$

I will be grateful if people with many flight hours can offer suggestions for $g(x,y)$. Needless to say, I am not asking that they verify those suggestions, but in case someone would like, these are the inputs to Wolfram integrator:

$\frac{1}{(g(x,y\text{as}r)}\sqrt{1-(kr/{x}^{(}1/3){)}^{2})}\phantom{\rule{0ex}{0ex}}\frac{(3t/4)(k/{t}^{1/3}{)}^{2}(2x)}{(g(x\text{as}t,y\text{as}x)\sqrt{1-(kx/{t}^{(}1/3){)}^{2})}}$

Differential EquationsAnswered question

Tiffany Page 2022-11-19

How to solve $x{y}^{\prime}=2\sqrt{{x}^{2}+{y}^{2}}+y$?

And what would be the standard form to illustrate this situation? (e.g. ${y}^{\prime}+P(x)y=Q(x)$ would be standard form of first order linear differential equation)

And what would be the standard form to illustrate this situation? (e.g. ${y}^{\prime}+P(x)y=Q(x)$ would be standard form of first order linear differential equation)

Differential EquationsAnswered question

klasyvea 2022-11-19

Separable Differential Equation dy/dt = 6y

The question is as follows:

$\frac{dy}{dt}=6y$

$y(9)=5$

I tried rearranging the equation to $\frac{dy}{6y}=dt$ and integrating both sides to get $(1/6)\mathrm{ln}|y|+C=t$. After that I tried plugging in the 9 for y and 5 for t and solving but I can't quite seem to get it.

The question is as follows:

$\frac{dy}{dt}=6y$

$y(9)=5$

I tried rearranging the equation to $\frac{dy}{6y}=dt$ and integrating both sides to get $(1/6)\mathrm{ln}|y|+C=t$. After that I tried plugging in the 9 for y and 5 for t and solving but I can't quite seem to get it.

Differential EquationsAnswered question

InjegoIrrenia1mk 2022-11-18

Is there a solution for y for $\frac{dy}{dx}=ax{e}^{by}$

I have come up with the equation in the form

$\frac{dy}{dx}=ax{e}^{by}$

, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or $\frac{{d}^{2}y}{d{x}^{2}}$?

I have come up with the equation in the form

$\frac{dy}{dx}=ax{e}^{by}$

, where a and b are arbitrary real numbers, for a project I am working on. I want to be able to find its integral and differentiation if possible. Does anyone know of a possible solution for $y$ and/or $\frac{{d}^{2}y}{d{x}^{2}}$?

Differential EquationsAnswered question

Hayley Mcclain 2022-11-18

All alternative solution for an equation

I'm looking for all alternative solutions of this

${x}^{\prime}=x(x-1)(x+1)$

But I absolutely don't know what I have to do! Thanks!

I'm looking for all alternative solutions of this

${x}^{\prime}=x(x-1)(x+1)$

But I absolutely don't know what I have to do! Thanks!

Differential EquationsAnswered question

Hanna Webster 2022-11-18

Separable Differential Equation explained

So i have the equation $\frac{dy}{dt}=1+y$

looking at the answers my lecture has given it states

$\frac{d}{dt}(1+y)=1+y$

then $1+y(t)=A{e}^{t}$ where A is a constant

Can someone explain these steps please?

So i have the equation $\frac{dy}{dt}=1+y$

looking at the answers my lecture has given it states

$\frac{d}{dt}(1+y)=1+y$

then $1+y(t)=A{e}^{t}$ where A is a constant

Can someone explain these steps please?

Differential EquationsAnswered question

klesstilne1 2022-11-17

How do I find the singular solution of the differential equation ${y}^{\prime}=\frac{{y}^{2}+1}{xy+y}$?

I start out with the separable differential equation,

${y}^{\prime}=\frac{dy}{dx}=\frac{{y}^{2}+1}{xy+y}=\frac{{y}^{2}+1}{y(x+1)}$

Thus, $\frac{1}{x+1}dx=\frac{y}{{y}^{2}+1}dy$

Then integrating both sides of the equation, I get

$\mathrm{ln}(x+1)=\frac{1}{2}\mathrm{ln}({y}^{2}+1)+C$

Now, ${e}^{\mathrm{ln}(x+1)}$ = ${e}^{\frac{1}{2}\mathrm{ln}({y}^{2}+1)+C}$. So...

$(x+1)={e}^{C}({y}^{2}+1{)}^{\frac{1}{2}}$

I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.

I start out with the separable differential equation,

${y}^{\prime}=\frac{dy}{dx}=\frac{{y}^{2}+1}{xy+y}=\frac{{y}^{2}+1}{y(x+1)}$

Thus, $\frac{1}{x+1}dx=\frac{y}{{y}^{2}+1}dy$

Then integrating both sides of the equation, I get

$\mathrm{ln}(x+1)=\frac{1}{2}\mathrm{ln}({y}^{2}+1)+C$

Now, ${e}^{\mathrm{ln}(x+1)}$ = ${e}^{\frac{1}{2}\mathrm{ln}({y}^{2}+1)+C}$. So...

$(x+1)={e}^{C}({y}^{2}+1{)}^{\frac{1}{2}}$

I kind of wanted to know if this is indeed the correct general formula. And also, how would I find the singular solution, if there happens to be one in this case.

Differential EquationsAnswered question

Arendrogfkl 2022-11-16

Solving separable equations with dy/dx on both sides?

I'm unsure on how to even start with this equation:

$y-x\frac{dy}{dx}=1+{x}^{2}\frac{dy}{dx}$

I thought it would be possible to cancel out but the answer was different. Could someone please provide a hint on how to start this equation? I thought about moving the $-x\frac{dy}{dx}$ over to the RHS and then group it under the common variable $\frac{dy}{dx}$ but the question was under the 'separable differential' section in my textbook so I don't think it's the correct method.

I'm unsure on how to even start with this equation:

$y-x\frac{dy}{dx}=1+{x}^{2}\frac{dy}{dx}$

I thought it would be possible to cancel out but the answer was different. Could someone please provide a hint on how to start this equation? I thought about moving the $-x\frac{dy}{dx}$ over to the RHS and then group it under the common variable $\frac{dy}{dx}$ but the question was under the 'separable differential' section in my textbook so I don't think it's the correct method.

Differential EquationsAnswered question

kunguwaat81 2022-11-15

Separable differential equation ${x}^{2}{y}^{\u2033}=2y$

Show that ${x}^{2}-{x}^{-1}$ is a solution of

${x}^{2}\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=2y$

I've tried separating the x and y terms and then integrating, but I could tell it wouldn't work. Any ideas? It's a simple question but I have not attempted them in a while.

Show that ${x}^{2}-{x}^{-1}$ is a solution of

${x}^{2}\frac{{\text{d}}^{2}y}{\text{d}{x}^{2}}=2y$

I've tried separating the x and y terms and then integrating, but I could tell it wouldn't work. Any ideas? It's a simple question but I have not attempted them in a while.

Differential EquationsAnswered question

Davirnoilc 2022-11-15

a separable differential equation

given this

$\frac{d}{dx}=x(1-x)$

where

$x(0)=0.1$

is this correct:?

$\frac{dx}{dt}=x(1-x)$

$t-{t}_{0}=\int \frac{dx}{x(1-x)}$

Where did the t and ${t}_{0}$ come from?

given this

$\frac{d}{dx}=x(1-x)$

where

$x(0)=0.1$

is this correct:?

$\frac{dx}{dt}=x(1-x)$

$t-{t}_{0}=\int \frac{dx}{x(1-x)}$

Where did the t and ${t}_{0}$ come from?

Differential EquationsAnswered question

Messiah Sutton 2022-11-15

First order non linear differential equation , non separable

First time I run into an equation of this "form", for non linear equations I know separation of variables and substitution. It doesn't seem to be homogenous of degree $0$ so the substitution $v=\frac{x}{t}$ is out of the question? Applying $\mathrm{ln}$ to both sides didn't do much either

${x}^{\prime}(t)={e}^{t+x(t)}-1,\phantom{\rule{1em}{0ex}}x(0)=1$

Wolfram gives the solution

$x(t)=-\mathrm{ln}({c}_{1}-t)-t$

First time I run into an equation of this "form", for non linear equations I know separation of variables and substitution. It doesn't seem to be homogenous of degree $0$ so the substitution $v=\frac{x}{t}$ is out of the question? Applying $\mathrm{ln}$ to both sides didn't do much either

${x}^{\prime}(t)={e}^{t+x(t)}-1,\phantom{\rule{1em}{0ex}}x(0)=1$

Wolfram gives the solution

$x(t)=-\mathrm{ln}({c}_{1}-t)-t$

Differential EquationsAnswered question

Sophie Marks 2022-11-14

How do you solve the following separable differential equation: y'y = y + 1?

I just started learning about differential equations and encountered following equation:

${y}^{\prime}y=y+1$

But I'm not sure how the integration is performed. What rules are used? How does $\int \frac{{y}^{\prime}y}{y+1}dx$ become $-\mathrm{log}(y+1)+y$?

I just started learning about differential equations and encountered following equation:

${y}^{\prime}y=y+1$

But I'm not sure how the integration is performed. What rules are used? How does $\int \frac{{y}^{\prime}y}{y+1}dx$ become $-\mathrm{log}(y+1)+y$?

Differential EquationsAnswered question

nyle2k8431 2022-11-13

2 separable O.D.Es.

a) ${y}^{\prime}sinx+ycosx=2cosx$

b) $\frac{1+{y}^{\prime}}{1-{y}^{\prime}}=\frac{1-\frac{y}{x}}{1+\frac{y}{x}}$

These are separable first order differential equations but i don't know where to start. I've tried substituting for u but it doesn't seem to work out. Thanks

a) ${y}^{\prime}sinx+ycosx=2cosx$

b) $\frac{1+{y}^{\prime}}{1-{y}^{\prime}}=\frac{1-\frac{y}{x}}{1+\frac{y}{x}}$

These are separable first order differential equations but i don't know where to start. I've tried substituting for u but it doesn't seem to work out. Thanks

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.