Solve differential equation

Kye
2021-02-23
Answered

Solve differential equation

You can still ask an expert for help

Alannej

Answered 2021-02-24
Author has **104** answers

That is

So

Integrating factor is

Now apply the initial condition y(0)=6 in the general solution

c=6-4=2 Now substitute 2 for c in general solution

Thus,the particular solution is

asked 2021-05-22

Write the following first-order differential equations in standard form. $\frac{dy}{dt}=yx(x+1)$

asked 2022-09-28

Find the general solution of $\left(\frac{dy}{dx}\right)={y}^{2}-4$ ?

asked 2021-02-21

Solve differential equation${y}^{\prime}+3{x}^{2}y=\mathrm{sin}\left(x\right){e}^{-{x}^{3}},\text{}y\left(0\right)=1$

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

that I have simplified like so

but I do not know how to solve this further to obtain the general solution. I have done first order linear differential equation strategies so far. How should I get about doing this question with the strategies I have?

asked 2022-07-01

The differential equation that describes my system is given as

${y}^{(n)}+{a}_{n-1}{y}^{(n-1)}+\cdots +{a}_{1}\dot{y}+{a}_{0}y={b}_{n-1}{u}^{(n-1)}+\cdots +{b}_{1}\dot{u}+{b}_{0}u+g(y(t),u(t))$

I want to express the above differential equation into a system of linear differential equations of the form

$\dot{x}=Ax+Bu+{B}_{p}g$

$y=Cx$

The matrices are given as follows: However, I am not able to prove, how to get them

$A=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ -{a}_{0}& -{a}_{1}& -{a}_{2}& \cdots & -{a}_{n-1}\end{array}\right]$

$C=\left[\begin{array}{ccccc}1& {b}_{1}/{b}_{0}& {b}_{2}/{b}_{0}& \cdots & {b}_{n-1}/{b}_{0}\end{array}\right]$

How do I get the above matrices from the differential equation form as shown above?

${y}^{(n)}+{a}_{n-1}{y}^{(n-1)}+\cdots +{a}_{1}\dot{y}+{a}_{0}y={b}_{n-1}{u}^{(n-1)}+\cdots +{b}_{1}\dot{u}+{b}_{0}u+g(y(t),u(t))$

I want to express the above differential equation into a system of linear differential equations of the form

$\dot{x}=Ax+Bu+{B}_{p}g$

$y=Cx$

The matrices are given as follows: However, I am not able to prove, how to get them

$A=\left[\begin{array}{ccccc}0& 1& 0& \cdots & 0\\ 0& 0& 1& \cdots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ -{a}_{0}& -{a}_{1}& -{a}_{2}& \cdots & -{a}_{n-1}\end{array}\right]$

$C=\left[\begin{array}{ccccc}1& {b}_{1}/{b}_{0}& {b}_{2}/{b}_{0}& \cdots & {b}_{n-1}/{b}_{0}\end{array}\right]$

How do I get the above matrices from the differential equation form as shown above?

asked 2022-09-11

Solve $\frac{dy}{dx}-\frac{1}{2}(1+\frac{1}{x})y+\frac{3}{x}{y}^{3}=0$?