How do you find all solutions of the differential equation $\frac{{d}^{2}y}{{dx}^{2}}=0$?

calcific5z
2022-09-11
Answered

How do you find all solutions of the differential equation $\frac{{d}^{2}y}{{dx}^{2}}=0$?

You can still ask an expert for help

Peyton Cox

Answered 2022-09-12
Author has **18** answers

This is a Second Order homogeneous DE with constant coefficients, but separable so we can just integrate (twice):

$\frac{{d}^{2}y}{{dx}^{2}}=0$

Integrate:

$\frac{dy}{dx}=A$

Integrate again:

$y=Ax+B$

which is the General Solution

$\frac{{d}^{2}y}{{dx}^{2}}=0$

Integrate:

$\frac{dy}{dx}=A$

Integrate again:

$y=Ax+B$

which is the General Solution

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 2022-06-11

I would like to solve:

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

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

$x=-2t$

That is, we have: $x=-2dy/dx$. Thus, we obtain

$y=-\frac{{x}^{2}}{4}+C$

Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

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

In order to do so, we let ${y}^{\prime}=t$, and we assume x as a function of t. Now, we take derivative with respect to t from the differential equation, and obtain

$\frac{dy}{dt}-x-t\frac{dx}{dt}-2t=0$

By the chain rule, we have: $dy/dt=tdx/dt$. So, the above simplifies to

$x=-2t$

That is, we have: $x=-2dy/dx$. Thus, we obtain

$y=-\frac{{x}^{2}}{4}+C$

Now, if we want to verify the solution, it turns out that C must be zero, in other words, $y=-{x}^{2}/4$ satisfies the original differential equation.

I have two questions:

1) What happens to the integration constant? That is, what is the general solution of the differential equation?

2) If we try to solve this differential equation with Mathematica, we obtain

$y={C}_{1}x+{C}_{1}^{2}$,

which has a different form from the analytical approach. How can we also produce this result analytically?

asked 2022-02-16

I understand that it applies to second order and any other differential equation as long as they are Homogenous and Linear, but I cant

asked 2021-05-01

Solve differential equation$x{y}^{\prime}+2y=-{x}^{3}+x,\text{}y(1)=2$

asked 2022-01-19

Dont

asked 2022-02-16

Suppose we have

$\frac{dy}{dx}+f\left(x\right)y=r\left(x\right)$

and it has two solutions${y}_{1}\left(x\right)$ and ${y}_{2}\left(x\right)$ then how to prove that solution of differential equation

$\frac{dy}{dx}+f\left(x\right)y=2r\left(x\right)$

Will be${y}_{1}\left(x\right)+{y}_{2}\left(x\right)$ ? I think given differential equations is linear first order equation so its solution will be

$y.{e}^{\int f\left(x\right)dx}=\int r.{e}^{\int f\left(x\right)dx}dx$

now do I establish two solution as$y}_{1$ and $y}_{2$ out of this equation?

and it has two solutions

Will be

now do I establish two solution as

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?