What is a solution to the differential equation ${e}^{x}(y\prime +1)=1$?

joguejaseg
2022-09-23
Answered

What is a solution to the differential equation ${e}^{x}(y\prime +1)=1$?

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altaryjny94

Answered 2022-09-24
Author has **14** answers

you can tell just by looking that this is separable, so we separate it out

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

$y\prime +1={e}^{-x}$

$y\prime ={e}^{-x}-1$

integrate both sides

$\int y\prime dx=\int {e}^{-x}-1dx$

$y=-{e}^{-x}-x+C$

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

$y\prime +1={e}^{-x}$

$y\prime ={e}^{-x}-1$

integrate both sides

$\int y\prime dx=\int {e}^{-x}-1dx$

$y=-{e}^{-x}-x+C$

asked 2022-09-11

What is a solution to the differential equation $\frac{dy}{dx}=\frac{y}{x}$?

asked 2022-04-10

For every differentiable function $f:\mathbb{R}\to \mathbb{R}$, there is a function $g:\mathbb{R}\times \mathbb{R}\to \mathbb{R}$ such that $g(f(x),{f}^{\prime}(x))=0$ for every $x$ and for every differentiable function $h:\mathbb{R}\to \mathbb{R}$ holds that

being true that for every $x\in \mathbb{R}$, $g(h(x),{h}^{\prime}(x))=0$ and $h(0)=f(0)$ implies that $h(x)=f(x)$ for every $x\in \mathbb{R}$.

i.e every differentiable function $f$ is a solution to some first order differential equation that has translation symmetry.

being true that for every $x\in \mathbb{R}$, $g(h(x),{h}^{\prime}(x))=0$ and $h(0)=f(0)$ implies that $h(x)=f(x)$ for every $x\in \mathbb{R}$.

i.e every differentiable function $f$ is a solution to some first order differential equation that has translation symmetry.

asked 2022-02-16

I've tried many times to reach the solution of a first order differential equation (of the last equation) but unfortunately I couldn't. Could you please help me to know how did he get this solution.

The equation is:

$\frac{dX}{dt}={k}_{f}-({k}_{f}+{k}_{b})X$

The author assumed that

$X=0$ at $t=0$

Then, he said in his first publication the solution is given as:

$X=\frac{{k}_{f}}{({k}_{f}+{k}_{b})}\{1-\mathrm{exp}[-({k}_{f}+{k}_{b})t]\}$

I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.

In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:

$X=\frac{{k}_{f}}{({k}_{f}+{k}_{b})}\{1-\mathrm{exp}[-({k}_{f}+{k}_{b})t]\}+{X}_{0}\text{}\mathrm{exp}[-({k}_{f}+{k}_{b})t]$ and he said $X}_{0$ was taken to be zero in the first publication.

Can any of you help how did the author get this solution please?

The equation is:

The author assumed that

Then, he said in his first publication the solution is given as:

I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.

In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:

Can any of you help how did the author get this solution please?

asked 2022-09-29

Considering this first-order linear differential equation:

$\frac{dy}{dx}+2y=0$

Although I now know the correct general solution to be $y={c}_{1}{e}^{-2x}$, I cannot figure out where I am going wrong with this apparently fallacious reasoning:

$$\frac{dy}{dx}+2y=0$$

$$\int \frac{dy}{dx}dx=\int -2y\text{}dx$$

$$y=-2xy+{c}_{1}$$

$$y(2x+1)={c}_{1}$$

$$y=\frac{{c}_{1}}{(2x+1)}$$

$\frac{dy}{dx}+2y=0$

Although I now know the correct general solution to be $y={c}_{1}{e}^{-2x}$, I cannot figure out where I am going wrong with this apparently fallacious reasoning:

$$\frac{dy}{dx}+2y=0$$

$$\int \frac{dy}{dx}dx=\int -2y\text{}dx$$

$$y=-2xy+{c}_{1}$$

$$y(2x+1)={c}_{1}$$

$$y=\frac{{c}_{1}}{(2x+1)}$$

asked 2022-07-10

I have a first order linear differential equation (a variation on a draining mixing tank problem) with many constants, and want to separate variables to solve it.

$\frac{dy}{dt}={k}_{1}+{k}_{2}\frac{y}{{k}_{3}+{k}_{4}t}$

y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, ${k}_{1}$ through ${k}_{4}$.

Separation of variables is made difficult by ${k}_{1}$, and I've considered an integrating factor, but think I might be missing something simple.

$\frac{dy}{dt}={k}_{1}+{k}_{2}\frac{y}{{k}_{3}+{k}_{4}t}$

y is the amount of mass in the tank at time t, and for simplicity, I've reduced various terms to constants, ${k}_{1}$ through ${k}_{4}$.

Separation of variables is made difficult by ${k}_{1}$, and I've considered an integrating factor, but think I might be missing something simple.

asked 2022-09-08

What is the general solution of the differential equation?

$(3{x}^{2}+2y)dx+2xdy=0$

$(3{x}^{2}+2y)dx+2xdy=0$

asked 2022-09-17

Solve the differential equation $\frac{dy}{dt}=4\sqrt{yt}$ y(1)=6?