Paul Reilly
2022-09-12
Answered

What is the general solution of the differential equation $\frac{dy}{dx}+2y=0$?

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zagrebova1c

Answered 2022-09-13
Author has **10** answers

We have:

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

We can just rearrange as follows:

$\frac{dy}{dx}=-2y\Rightarrow \frac{1}{y}\frac{dy}{dx}=-2$

This is a first Order Separable Differential Equation and "separate the variables" to get

$\int \frac{1}{y}dy=\int -2dx$

And integrating gives us:

$\mathrm{ln}\left|y\right|=-2x+A$

$\therefore \left|y\right|={e}^{-2x+A}$

Note that as $e}^{\alpha}>0\forall \alpha \in \mathbb{R$, we can write

$y={e}^{-2x+A}$

$={e}^{-2x}{e}^{A}$

$=C{e}^{-2x}$

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

We can just rearrange as follows:

$\frac{dy}{dx}=-2y\Rightarrow \frac{1}{y}\frac{dy}{dx}=-2$

This is a first Order Separable Differential Equation and "separate the variables" to get

$\int \frac{1}{y}dy=\int -2dx$

And integrating gives us:

$\mathrm{ln}\left|y\right|=-2x+A$

$\therefore \left|y\right|={e}^{-2x+A}$

Note that as $e}^{\alpha}>0\forall \alpha \in \mathbb{R$, we can write

$y={e}^{-2x+A}$

$={e}^{-2x}{e}^{A}$

$=C{e}^{-2x}$

asked 2022-09-10

How to you find the general solution of $\frac{dy}{dx}=\mathrm{sin}2x$?

asked 2022-05-24

How to solve this differential equation:

$x\frac{dy}{dx}=y+x\frac{{e}^{x}}{{e}^{y}}?$

I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v=\frac{y}{x}$ to solve it.

$x\frac{dy}{dx}=y+x\frac{{e}^{x}}{{e}^{y}}?$

I tried to rearrange the equation to the form $f\left(\frac{y}{x}\right)$ but I couldn't thus I couldn't use $v=\frac{y}{x}$ to solve it.

asked 2020-11-14

Solve differential equation

asked 2022-06-21

Consider $x{y}^{\u2033}+2{y}^{\prime}+xy=0$. Its solutions are $\frac{\mathrm{cos}x}{x},\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{sin}x}{x}$.

Neither of those solutions (as far as I could find) can be the solutions of a first order linear homogeneous differential equation. However

$\frac{{e}^{\pm ix}}{x}$

Can be the solution of a first order linear homogeneous DE (${y}^{\prime}+(x\mp i)y=0$)

Can I always find a first order homogeneous linear DE whose solution also solves a second order homogeneous linear DE?

For example can I find a first order homogeneous linear DE whose solution is a particular linear combination of ${J}_{1}$ and ${Y}_{1}$?

(unrelated: also I'd like to know if there is a way of solving $x{y}^{\u2033}+2{y}^{\prime}+xy=0$ without noticing that it is a spherical bessel function or using laplace transform.)

Neither of those solutions (as far as I could find) can be the solutions of a first order linear homogeneous differential equation. However

$\frac{{e}^{\pm ix}}{x}$

Can be the solution of a first order linear homogeneous DE (${y}^{\prime}+(x\mp i)y=0$)

Can I always find a first order homogeneous linear DE whose solution also solves a second order homogeneous linear DE?

For example can I find a first order homogeneous linear DE whose solution is a particular linear combination of ${J}_{1}$ and ${Y}_{1}$?

(unrelated: also I'd like to know if there is a way of solving $x{y}^{\u2033}+2{y}^{\prime}+xy=0$ without noticing that it is a spherical bessel function or using laplace transform.)

asked 2022-01-22

Maybe you know how to do it?

Write an equivalent first-order differential equation and initial condition for y

What is the equivalent first-order differential equation?

asked 2022-01-21

Find the critical points and sketch the phase portrait of the given autonomous first order differential equation. Classify the each critical point as asmyptotically stable, unstable, or semi-stable.

${y}^{\prime}={y}^{2}(4-{y}^{2})$

asked 2022-06-18

I need to convert the following differential equation to standard form.

${T}_{n}=2{T}_{n-1}+1$

(not quite sure how to really format it properly)

I was thinking it is

${T}_{n}-2{T}_{n-1}-1$

If I'm incorrect (or correct) can I please have somebody just explain briefly about the decreasing order. I'm just confused about the whole number, in this case the number 1, I'm not quite sure if it's bigger or smaller than the others.

${T}_{n}=2{T}_{n-1}+1$

(not quite sure how to really format it properly)

I was thinking it is

${T}_{n}-2{T}_{n-1}-1$

If I'm incorrect (or correct) can I please have somebody just explain briefly about the decreasing order. I'm just confused about the whole number, in this case the number 1, I'm not quite sure if it's bigger or smaller than the others.