How do you find all solutions of the differential equation $\frac{{d}^{2}y}{{dx}^{2}}={x}^{-2}$?

frobirrimupyx
2022-09-13
Answered

How do you find all solutions of the differential equation $\frac{{d}^{2}y}{{dx}^{2}}={x}^{-2}$?

You can still ask an expert for help

asked 2022-04-10

Problem:

Solve the following differential equations.

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

Answer:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ \frac{dv}{dx}+3v& =3\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have P(x)=3.

$\begin{array}{rl}I(x)& ={e}^{3x}\\ {e}^{3x}\frac{dv}{dx}+3{e}^{3x}v& =3{e}^{3x}\\ D\left({e}^{3x}v\right)& =3{e}^{3x}\\ {e}^{3x}v& ={e}^{3x}+C\\ v& =1+{e}^{-3x}\\ {y}^{3}& =1+{e}^{-3x}\end{array}$

Now to check my answer.

$\begin{array}{rl}3{y}^{2}\frac{dy}{dx}& =-3C{e}^{-3x}\\ {y}^{2}\frac{dy}{dx}& =-C{e}^{-3x}\\ \frac{dy}{dx}& =-C{e}^{-3x}{y}^{-2}\\ x\frac{dy}{dx}+y& =x(-C{e}^{-3x}{y}^{-2})+{(1+{e}^{-3x})}^{\frac{1}{3}}\end{array}$

I cannot seem to get my answer to check. Where did I go wrong?

Here is my second attempt to solve the problem:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}$

Now, I want to write:

$D(3xv)=9$

but that is wrong. What did I do wrong?

Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& ={x}^{3}\\ {x}^{3}\frac{dv}{dx}+3{x}^{2}v& =3{x}^{2}\\ D({x}^{3}v)& ={x}^{3}+C\\ {x}^{3}v& ={x}^{3}+C\\ v& =C{x}^{-3}+1\\ {y}^{3}& =C{x}^{-3}+1\end{array}$

Do I have it right now?

Solve the following differential equations.

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

Answer:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ \frac{dv}{dx}+3v& =3\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have P(x)=3.

$\begin{array}{rl}I(x)& ={e}^{3x}\\ {e}^{3x}\frac{dv}{dx}+3{e}^{3x}v& =3{e}^{3x}\\ D\left({e}^{3x}v\right)& =3{e}^{3x}\\ {e}^{3x}v& ={e}^{3x}+C\\ v& =1+{e}^{-3x}\\ {y}^{3}& =1+{e}^{-3x}\end{array}$

Now to check my answer.

$\begin{array}{rl}3{y}^{2}\frac{dy}{dx}& =-3C{e}^{-3x}\\ {y}^{2}\frac{dy}{dx}& =-C{e}^{-3x}\\ \frac{dy}{dx}& =-C{e}^{-3x}{y}^{-2}\\ x\frac{dy}{dx}+y& =x(-C{e}^{-3x}{y}^{-2})+{(1+{e}^{-3x})}^{\frac{1}{3}}\end{array}$

I cannot seem to get my answer to check. Where did I go wrong?

Here is my second attempt to solve the problem:

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$.

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& =3x\\ 3x\frac{dv}{dx}+9v& =9\end{array}$

Now, I want to write:

$D(3xv)=9$

but that is wrong. What did I do wrong?

Here is my third attempt to solve the problem. Last time, I made a mistake in finding the integrating factor.

To solve this equation, we reduce it to a linear differential equation with the substitution $v={y}^{3}$

$\begin{array}{rl}x{y}^{2}\frac{dy}{dx}+{y}^{3}& =1\\ \frac{dv}{dx}& =3{y}^{2}\frac{dy}{dx}\\ 3x{y}^{2}\frac{dy}{dx}+3{y}^{3}& =3\\ x\frac{dv}{dx}+3v& =3\\ \frac{dv}{dx}+3{x}^{-1}v& =3{x}^{-1}\end{array}$

Now we have a first order linear differential equation. To solve it, we use the integrating factor $I={e}^{\int P(x)}$. In this case, we have $P(x)=3{x}^{-1}$.

$\begin{array}{rl}I& ={e}^{3\int {x}^{-1}\phantom{\rule{thinmathspace}{0ex}}dx}={e}^{3\mathrm{ln}|x|}\\ I& ={x}^{3}\\ {x}^{3}\frac{dv}{dx}+3{x}^{2}v& =3{x}^{2}\\ D({x}^{3}v)& ={x}^{3}+C\\ {x}^{3}v& ={x}^{3}+C\\ v& =C{x}^{-3}+1\\ {y}^{3}& =C{x}^{-3}+1\end{array}$

Do I have it right now?

asked 2022-01-19

Classify the equation as separable, linear, both, or neither.

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

asked 2022-02-15

I really dont know what to do whit this equation, i have tried to integrate it, but i can do it.

${y}^{\prime 3}-y{y}^{\prime 2}-{x}^{2}{y}^{\prime}+{x}^{2}y=0$

asked 2022-01-20

Help!

a) Solve the first order linear differential equation$\frac{dy}{dx}+xy=x,\text{}{y}_{0}=-6$ . b) Write down any two applications of first order linear differential equation.

a) Solve the first order linear differential equation

asked 2022-09-10

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asked 2021-03-07

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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.