Solve the differential equation $\frac{dy}{dx}+y=0$ ?

Aidyn Meza
2022-09-27
Answered

Solve the differential equation $\frac{dy}{dx}+y=0$ ?

You can still ask an expert for help

elilsonoulp2l

Answered 2022-09-28
Author has **9** answers

We can rewrite the equation:

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

as:

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

Which is a First Order linear separable Differential Equation, so we can "separate the variables" to get:

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

Which we can integrate to get:

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

Taking Natural logarithms we then get:

$\left|y\right|={e}^{-x+c}$

As $e}^{x}>0\forall x\in \mathbb{R$ we then get:

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

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

We can easily verify the solution:

$y=A{e}^{-x}\Rightarrow y\prime =-A{e}^{-x}$

$y\prime +y=-A{e}^{-x}+A{e}^{-x}=0QED$

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

as:

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

Which is a First Order linear separable Differential Equation, so we can "separate the variables" to get:

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

Which we can integrate to get:

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

Taking Natural logarithms we then get:

$\left|y\right|={e}^{-x+c}$

As $e}^{x}>0\forall x\in \mathbb{R$ we then get:

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

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

We can easily verify the solution:

$y=A{e}^{-x}\Rightarrow y\prime =-A{e}^{-x}$

$y\prime +y=-A{e}^{-x}+A{e}^{-x}=0QED$

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Solve the Differential equations

asked 2022-02-15

Throughout my engineering education, Ive

asked 2022-06-11

I would like some help on comprehending this question as well as a push in the right direction. The question gave a system of first-order differential equation.

$x(t{)}^{\prime}=4x(t)-3y(t)+6{e}^{2t}$

$y(t{)}^{\prime}=4x(t)-6y(t)$

The question asked me to find the 2nd inhomogeneous equation that satisfies x(t). Does this mean the answer should all be in terms of x? I tried focusing on the x and differentiating it with respect to t.

so $x(t{)}^{\prime}=4x(t)-3y(t{)}^{\prime}+6{e}^{2t}$ becomes:

$x(t{)}^{\u2033}=4x(t{)}^{\prime}-3y(t{)}^{\prime}+12{e}^{2t}$

for simplicity sake, I will write x(t) as x and y(t) as y.

After that step, I replaced the y' in the x'' equation with the rearranged y' from the original question into the differentiated x' equation. This gives:

${x}^{\u2033}=4{x}^{\prime}-3(4x-6(\frac{1}{-3}({x}^{\prime}-4x-6{e}^{2t})+12{e}^{2t}$

this cancels down to:

${x}^{\u2033}=4{x}^{\prime}-12x+2{x}^{\prime}-8x$

but if you move everything to one side, it becomes

${x}^{\u2033}-6{x}^{\prime}+20x=0$

this is a second-order homogenous equation, so I don't quite know where I went wrong

$x(t{)}^{\prime}=4x(t)-3y(t)+6{e}^{2t}$

$y(t{)}^{\prime}=4x(t)-6y(t)$

The question asked me to find the 2nd inhomogeneous equation that satisfies x(t). Does this mean the answer should all be in terms of x? I tried focusing on the x and differentiating it with respect to t.

so $x(t{)}^{\prime}=4x(t)-3y(t{)}^{\prime}+6{e}^{2t}$ becomes:

$x(t{)}^{\u2033}=4x(t{)}^{\prime}-3y(t{)}^{\prime}+12{e}^{2t}$

for simplicity sake, I will write x(t) as x and y(t) as y.

After that step, I replaced the y' in the x'' equation with the rearranged y' from the original question into the differentiated x' equation. This gives:

${x}^{\u2033}=4{x}^{\prime}-3(4x-6(\frac{1}{-3}({x}^{\prime}-4x-6{e}^{2t})+12{e}^{2t}$

this cancels down to:

${x}^{\u2033}=4{x}^{\prime}-12x+2{x}^{\prime}-8x$

but if you move everything to one side, it becomes

${x}^{\u2033}-6{x}^{\prime}+20x=0$

this is a second-order homogenous equation, so I don't quite know where I went wrong

asked 2022-05-15

I was wondering if I could get some advice on how to tackle this question:

Consider the differential equation

${x}^{2}\frac{dy}{dx}+2xy-{y}^{3}=0\phantom{\rule{1em}{0ex}}(3)$

Make the substitution $u={y}^{-2}$ and show that the differential equation reduces to

$-\frac{1}{2}{x}^{2}\frac{du}{dx}+2xu-1=0\phantom{\rule{1em}{0ex}}(4)$

Solve equation (4) for u(x) and hence write down the solution for equation (3).

I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:

$\begin{array}{rl}u& ={y}^{-2}\\ & =\frac{1}{{y}^{2}}\\ \therefore {y}^{2}& =\frac{1}{u}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y& =\pm \sqrt{\frac{1}{u}}\end{array}$

I'm not sure where to continue on from here though.

Consider the differential equation

${x}^{2}\frac{dy}{dx}+2xy-{y}^{3}=0\phantom{\rule{1em}{0ex}}(3)$

Make the substitution $u={y}^{-2}$ and show that the differential equation reduces to

$-\frac{1}{2}{x}^{2}\frac{du}{dx}+2xu-1=0\phantom{\rule{1em}{0ex}}(4)$

Solve equation (4) for u(x) and hence write down the solution for equation (3).

I'm trying to do the first part of showing that the differential equation reduces to equation 4. I have started out by:

$\begin{array}{rl}u& ={y}^{-2}\\ & =\frac{1}{{y}^{2}}\\ \therefore {y}^{2}& =\frac{1}{u}\\ \phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y& =\pm \sqrt{\frac{1}{u}}\end{array}$

I'm not sure where to continue on from here though.

asked 2022-10-27

${x}^{2}{y}^{\u2033}+x{y}^{\prime}-y={x}^{2}$

My attempt:

Divided by ${x}^{2}$:

${y}^{\u2033}+\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}=1$

Now to solve the homogenous equation using Euler's method

${y}^{\u2033}+\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}=0$

To look for solution $y={x}^{r}$

so ${y}^{\prime}=r{x}^{r-1}$

${y}^{\u2033}=r(r-1){x}^{r-2}$

So:

$r(r-1){x}^{r-2}+\frac{r{x}^{r-1}}{x}-\frac{{x}^{r}}{{x}^{2}}=0$

Divided by ${x}^{r}$:

$r(r-1){x}^{-2}+\frac{r{x}^{-1}}{x}-\frac{1}{{x}^{2}}=0$

Is it correct so far?

My problem: I don't know how to find ${r}_{1},{r}_{2}$

My attempt:

Divided by ${x}^{2}$:

${y}^{\u2033}+\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}=1$

Now to solve the homogenous equation using Euler's method

${y}^{\u2033}+\frac{{y}^{\prime}}{x}-\frac{y}{{x}^{2}}=0$

To look for solution $y={x}^{r}$

so ${y}^{\prime}=r{x}^{r-1}$

${y}^{\u2033}=r(r-1){x}^{r-2}$

So:

$r(r-1){x}^{r-2}+\frac{r{x}^{r-1}}{x}-\frac{{x}^{r}}{{x}^{2}}=0$

Divided by ${x}^{r}$:

$r(r-1){x}^{-2}+\frac{r{x}^{-1}}{x}-\frac{1}{{x}^{2}}=0$

Is it correct so far?

My problem: I don't know how to find ${r}_{1},{r}_{2}$

asked 2021-01-16

Find the general solution of the first-order linear differential equation
$(dy/dx)+(1/x)y=6x+2$ , for x > 0