How do you solve for xy'−y=3xy given y(1)=0?

overrated3245w
2022-09-29
Answered

How do you solve for xy'−y=3xy given y(1)=0?

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Matteo Estes

Answered 2022-09-30
Author has **9** answers

$x\frac{dy}{dx}=3xy+y$

$x\frac{dy}{dx}=y(3x+1)$

$\frac{dy}{y}=\frac{3x+1}{x}dx$

Integrating;

$\int \frac{dy}{y}=\int \frac{3x+1}{x}dx$

$\int \frac{dy}{y}=\int (3+\frac{1}{x})dx$

$\mathrm{ln}\left(y\right)=3x+\mathrm{ln}\left(x\right)+c$

$y={e}^{3x+\mathrm{ln}\left(x\right)+c}$

And finally;

$y={e}^{3x}\times {e}^{\mathrm{ln}\left(x\right)}\times {e}^{c}=Ax{e}^{3x}$

where $A={e}^{c}$

We can now find the value of the constant A;

$0=A{e}^{3}$

A=0

The particular solution to your equation is then:

$y=0x{e}^{3x}=0$

$x\frac{dy}{dx}=y(3x+1)$

$\frac{dy}{y}=\frac{3x+1}{x}dx$

Integrating;

$\int \frac{dy}{y}=\int \frac{3x+1}{x}dx$

$\int \frac{dy}{y}=\int (3+\frac{1}{x})dx$

$\mathrm{ln}\left(y\right)=3x+\mathrm{ln}\left(x\right)+c$

$y={e}^{3x+\mathrm{ln}\left(x\right)+c}$

And finally;

$y={e}^{3x}\times {e}^{\mathrm{ln}\left(x\right)}\times {e}^{c}=Ax{e}^{3x}$

where $A={e}^{c}$

We can now find the value of the constant A;

$0=A{e}^{3}$

A=0

The particular solution to your equation is then:

$y=0x{e}^{3x}=0$

asked 2022-02-15

Most systems can be modeled using first order system just like a temperature control system of an incubator.

Question 1: How can we know that a certain system can be modeled using first order differential equation?

${a}_{1}\frac{dy}{dt}+{a}_{0}y=F\left(t\right)\text{}\text{}\text{}\text{}\left(1\right)$

A first order differential equation in Equation (1) may be written as:

$\tau \frac{dy}{dt}+y=kF\left(t\right)\text{}\text{}\text{}\text{}\left(2\right)$

Question 2: If a step input

$F\left(t\right)=\{\begin{array}{cc}0& \text{}\text{}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t0\\ A& \text{}\text{}\text{}{\textstyle \phantom{\rule{1em}{0ex}}}\text{if}{\textstyle \phantom{\rule{1em}{0ex}}}\text{}t0\end{array}\text{}\text{}\text{}\text{}\left(3\right)$

of the system has an output response of a

$y\left(t\right)=kA(1-{e}^{\frac{-t}{\tau}}\text{}\text{}\text{}\text{}\left(4\right)$

then the system can be modeled as first differential equation where

${y}_{max}=kA\text{}\text{}\text{}\text{}\left(5\right)$

Is this correct?

Given a step input and the graph of the step response of the system. Am i correct if I say that if the output of the system behaves similar to equation (4), then I can model the system using first differential equation?

Question 1: How can we know that a certain system can be modeled using first order differential equation?

A first order differential equation in Equation (1) may be written as:

Question 2: If a step input

of the system has an output response of a

then the system can be modeled as first differential equation where

Is this correct?

Given a step input and the graph of the step response of the system. Am i correct if I say that if the output of the system behaves similar to equation (4), then I can model the system using first differential equation?

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.

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In a course on partial differential equations I came through this theorem about the general solution of a first order quasi-linear partial differential equation.

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

1. The general solution of a first-order, quasi-linear partial differential equation

$a(x,y,u){u}_{x}+b(x,y,u){u}_{y}=c(x,y,u)$

is given by $f(\varphi ,\psi )=0$, where $f$ is an arbitrary function of $\varphi (x,y,u)$ and $\psi (x,y,u).$.

2. $\varphi =C1$ and $\psi =C2$ are solution curves of the characteristic equations

$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$

Is there any geometric interpretation of both these points so that I can have a better intuitive understanding of the graphical representation of $f$,$\varphi $ and $\psi $ ?

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What is a solution to the differential equation $\frac{dy}{dx}=\frac{\mathrm{sin}x}{y}$?

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What is the solution to the Differential Equation $\frac{4}{{y}^{3}}\frac{dy}{dx}=\frac{1}{x}$?