Dtjytyk

Is having access to past exams cheating and, if yes, could it be proven just by a good grade?

Is it normal that my co-workers at a fitness company criticize my food choices?

Meaning of "SEVERA INDEOVI VAS" from 3rd Century slab

Bash: What does "masking return values" mean?

Calculus II Professor will not accept my correct integral evaluation that uses a different method, should I bring this up further?

Can elves maintain concentration in a trance?

Why doesn't the EU now just force the UK to choose between referendum and no-deal?

Why are there 40 737 Max planes in flight when they have been grounded as not airworthy?

How to deal with a cynical class?

Be in awe of my brilliance!

Latest web browser compatible with Windows 98

What are some nice/clever ways to introduce the tonic's dominant seventh chord?

It's a yearly task, alright

The use of "touch" and "touch on" in context

I need to drive a 7/16" nut but am unsure how to use the socket I bought for my screwdriver

Should we release the security issues we found in our product as CVE or we can just update those on weekly release notes?

Russian cases: A few examples, I'm really confused

Is a lawful good "antagonist" effective?

Rules about breaking the rules. How do I do it well?

How to answer questions about my characters?

What is the greatest age difference between a married couple in Tanach?

Why do passenger jet manufacturers design their planes with stall prevention systems?

Is it true that real estate prices mainly go up?

Where is the 1/8 CR apprentice in Volo's Guide to Monsters?

What are differential equations and how do you solve ${dy over dx}=y$ and find $y$ in terms $x$?

How do you solve differential equations in the form of $ay'' + by' +cy = d$?How do you solve second-order linear differential equations without inspection?how to solve second order differential equationFind the maximum area of a right triangle with a constant perimeter P.How do you know when to use the Chain Rule instead of the Power Rule?What method should I apply to solve this differential equation?How to solve differential equations that look like theseSolve System of Differential equations w/o using matrices.How to solve differential equation with one differential termFind the integrating factor and solve

-1

1

$begingroup$

I had been wondering about how to solve the equation $${dy over dx}=y$$. My progress was to use the chain rule, like setting $$z=2x;{dyover dx}={dyover dz}*{dzover dx}={dyover d(2x)}*2=y.$$ Now I’m officially stuck. What am I supposed to do?
An answer that do work(just for checking): $e^{x+1}$

ordinary-differential-equations derivatives exponential-function chain-rule

share|cite|improve this question

asked Mar 10 at 14:18

Math LoverMath Lover

16010

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e$^x$ also works
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $int {dy over y} = int dx$
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it's variable separable ODE also linear .
  $endgroup$
  – Faraday Pathak
  Mar 10 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner I know that.
  $endgroup$
  – Math Lover
  Mar 10 at 23:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  THANKS A LOT FOR YOUR ANSWERS!
  $endgroup$
  – Math Lover
  Mar 10 at 23:50
  

  
  
  
  

add a comment | 

-1

1

$begingroup$

I had been wondering about how to solve the equation $${dy over dx}=y$$. My progress was to use the chain rule, like setting $$z=2x;{dyover dx}={dyover dz}*{dzover dx}={dyover d(2x)}*2=y.$$ Now I’m officially stuck. What am I supposed to do?
An answer that do work(just for checking): $e^{x+1}$

ordinary-differential-equations derivatives exponential-function chain-rule

share|cite|improve this question

asked Mar 10 at 14:18

Math LoverMath Lover

16010

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  e$^x$ also works
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:20
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $int {dy over y} = int dx$
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  it's variable separable ODE also linear .
  $endgroup$
  – Faraday Pathak
  Mar 10 at 14:26
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner I know that.
  $endgroup$
  – Math Lover
  Mar 10 at 23:46
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  THANKS A LOT FOR YOUR ANSWERS!
  $endgroup$
  – Math Lover
  Mar 10 at 23:50
  

  
  
  
  

add a comment | 

$begingroup$

I had been wondering about how to solve the equation $${dy over dx}=y$$. My progress was to use the chain rule, like setting $$z=2x;{dyover dx}={dyover dz}*{dzover dx}={dyover d(2x)}*2=y.$$ Now I’m officially stuck. What am I supposed to do?
An answer that do work(just for checking): $e^{x+1}$

ordinary-differential-equations derivatives exponential-function chain-rule

share|cite|improve this question

asked Mar 10 at 14:18

Math LoverMath Lover

16010

$endgroup$

share|cite|improve this question

asked Mar 10 at 14:18

Math LoverMath Lover

16010

share|cite|improve this question

asked Mar 10 at 14:18

Math LoverMath Lover

16010

asked Mar 10 at 14:18

Math LoverMath Lover

16010

asked Mar 10 at 14:18

Math LoverMath Lover

16010

4 Answers
4

active

oldest

votes

1

$begingroup$

This is a separable differential equation. Assuming $yne0$ which is a condition we will check separately, divide both sides of the equation by $y$:

$$
frac{dy}{dx}=y\
frac{1}{y}frac{dy}{dx}=1\
intfrac{1}{y}frac{dy}{dx},dx=int,dx\
intfrac{1}{y},dy=x+C_1\
ln{|y|}+C_2=x+C_1\
ln{|y|}=x+C_1-C_2\
ln{|y|}=x+C_3\
|y|=e^{x+C_3}\
y=pm e^{C_3}e^{x}
$$

$pm e^{C_3}$ covers all real numbers except $0$. But $y=0$ is also a solution because $0'=0$. Therefore, we can write the following as the general solution to this equation:

$$y=Ce^x, Cinmathbb{R}.$$

share|cite|improve this answer

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

$endgroup$

add a comment | 

4

$begingroup$

$frac{dy}{dx}=yRightarrowfrac{dy}{y}=dxRightarrowintfrac{dy}{y}=int dx$

$Rightarrow lnleft(yright)=x+CRightarrow y=e^{x+C}=C'e^x$

$C$ and $C'=e^{C}$ are constants; in order to put integrals, you should quote the theorem of separation of variables for differential equations.

share|cite|improve this answer

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected $C’e$ to $C’e^x$
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My mistake typing it, thank you!
  $endgroup$
  – Jack Talion
  Mar 10 at 14:46
  

  
  
  
  

add a comment | 

1

$begingroup$

Here comes a solution without "magical" manipulation of differentials.

Assuming that $y neq 0$ on some interval, divide the equation with $y$:
$$frac{y'(x)}{y(x)} = 1.$$

Now the left hand side can be written as the derivative of $ln |y(x)|$ and the right hand side as the derivative of $x$:
$$frac{d}{dx} ln |y(x)| = frac{d}{dx} x.$$

Therefore, $ln |y(x)| = x + C,$ where $C$ is some constant, and on the interval we thus have
$$y(x) = pm e^{x+C} = pm e^C e^x = C' e^x,$$
where $C' = pm e^C$ is a nonzero constant.

We can see though that also when $C' = 0,$ we get a solution: $y(x) equiv 0.$ We therefore have a general solution $y(x) = C' e^x,$ where $C'$ is a constant.

share|cite|improve this answer

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

1

$begingroup$

The first question in your title is

What are differential equations?

The informal answer is that often what we know about a quantity we are interested in is its rate of change. For example, if you know the velocity as a function $v(t)$ of time you can find the position $p$ by solving the differential equation
$$
p'(t) = v(t) .
$$
That's just integration. You use any known position to find the "constant of integration".

The differential equation you ask about,
$$
y' = y,
$$
says that the rate of change of $y$ is $y$ itself. In slightly more generality
$$
y' = ky
$$
says that the rate of change of $y$ is proportional to $y$. That classic equation models population growth over time or compound interest: the more stuff there is the faster it's growing. The solution is exponential growth:
$$
y = Ce^{kt}.
$$

The second order differential equation
$$
y" = -ky
$$
models a spring: the second derivative of position is (essentially) a force (that's Newton's law $F=ma$). The equation says that the farther $y$ is from $0$ the greater the force returning it to $0$. The solutions are the oscillating functions $y = sin(t+ phi)$.

share|cite|improve this answer

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

$endgroup$

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3142424%2fwhat-are-differential-equations-and-how-do-you-solve-dy-over-dx-y-and-find%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

This is a separable differential equation. Assuming $yne0$ which is a condition we will check separately, divide both sides of the equation by $y$:

$$
frac{dy}{dx}=y\
frac{1}{y}frac{dy}{dx}=1\
intfrac{1}{y}frac{dy}{dx},dx=int,dx\
intfrac{1}{y},dy=x+C_1\
ln{|y|}+C_2=x+C_1\
ln{|y|}=x+C_1-C_2\
ln{|y|}=x+C_3\
|y|=e^{x+C_3}\
y=pm e^{C_3}e^{x}
$$

$pm e^{C_3}$ covers all real numbers except $0$. But $y=0$ is also a solution because $0'=0$. Therefore, we can write the following as the general solution to this equation:

$$y=Ce^x, Cinmathbb{R}.$$

share|cite|improve this answer

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

$endgroup$

add a comment | 

1

$begingroup$

This is a separable differential equation. Assuming $yne0$ which is a condition we will check separately, divide both sides of the equation by $y$:

$$
frac{dy}{dx}=y\
frac{1}{y}frac{dy}{dx}=1\
intfrac{1}{y}frac{dy}{dx},dx=int,dx\
intfrac{1}{y},dy=x+C_1\
ln{|y|}+C_2=x+C_1\
ln{|y|}=x+C_1-C_2\
ln{|y|}=x+C_3\
|y|=e^{x+C_3}\
y=pm e^{C_3}e^{x}
$$

$pm e^{C_3}$ covers all real numbers except $0$. But $y=0$ is also a solution because $0'=0$. Therefore, we can write the following as the general solution to this equation:

$$y=Ce^x, Cinmathbb{R}.$$

share|cite|improve this answer

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

$endgroup$

add a comment | 

$begingroup$

This is a separable differential equation. Assuming $yne0$ which is a condition we will check separately, divide both sides of the equation by $y$:

$$
frac{dy}{dx}=y\
frac{1}{y}frac{dy}{dx}=1\
intfrac{1}{y}frac{dy}{dx},dx=int,dx\
intfrac{1}{y},dy=x+C_1\
ln{|y|}+C_2=x+C_1\
ln{|y|}=x+C_1-C_2\
ln{|y|}=x+C_3\
|y|=e^{x+C_3}\
y=pm e^{C_3}e^{x}
$$

$pm e^{C_3}$ covers all real numbers except $0$. But $y=0$ is also a solution because $0'=0$. Therefore, we can write the following as the general solution to this equation:

$$y=Ce^x, Cinmathbb{R}.$$

share|cite|improve this answer

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

$endgroup$

share|cite|improve this answer

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

edited Mar 11 at 1:08

J. W. Tanner

3,2401320

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

answered Mar 10 at 16:40

Michael RybkinMichael Rybkin

3,884420

4

$begingroup$

$frac{dy}{dx}=yRightarrowfrac{dy}{y}=dxRightarrowintfrac{dy}{y}=int dx$

$Rightarrow lnleft(yright)=x+CRightarrow y=e^{x+C}=C'e^x$

$C$ and $C'=e^{C}$ are constants; in order to put integrals, you should quote the theorem of separation of variables for differential equations.

share|cite|improve this answer

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected $C’e$ to $C’e^x$
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My mistake typing it, thank you!
  $endgroup$
  – Jack Talion
  Mar 10 at 14:46
  

  
  
  
  

add a comment | 

4

$begingroup$

$frac{dy}{dx}=yRightarrowfrac{dy}{y}=dxRightarrowintfrac{dy}{y}=int dx$

$Rightarrow lnleft(yright)=x+CRightarrow y=e^{x+C}=C'e^x$

$C$ and $C'=e^{C}$ are constants; in order to put integrals, you should quote the theorem of separation of variables for differential equations.

share|cite|improve this answer

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I corrected $C’e$ to $C’e^x$
  $endgroup$
  – J. W. Tanner
  Mar 10 at 14:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My mistake typing it, thank you!
  $endgroup$
  – Jack Talion
  Mar 10 at 14:46
  

  
  
  
  

add a comment | 

$begingroup$

$frac{dy}{dx}=yRightarrowfrac{dy}{y}=dxRightarrowintfrac{dy}{y}=int dx$

$Rightarrow lnleft(yright)=x+CRightarrow y=e^{x+C}=C'e^x$

$C$ and $C'=e^{C}$ are constants; in order to put integrals, you should quote the theorem of separation of variables for differential equations.

share|cite|improve this answer

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

edited Mar 10 at 16:24

J. W. Tanner

3,2401320

answered Mar 10 at 14:26

Jack TalionJack Talion

886

New contributor

Jack Talion is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Mar 10 at 14:26

Jack TalionJack Talion

886

1

$begingroup$

Here comes a solution without "magical" manipulation of differentials.

Assuming that $y neq 0$ on some interval, divide the equation with $y$:
$$frac{y'(x)}{y(x)} = 1.$$

Now the left hand side can be written as the derivative of $ln |y(x)|$ and the right hand side as the derivative of $x$:
$$frac{d}{dx} ln |y(x)| = frac{d}{dx} x.$$

Therefore, $ln |y(x)| = x + C,$ where $C$ is some constant, and on the interval we thus have
$$y(x) = pm e^{x+C} = pm e^C e^x = C' e^x,$$
where $C' = pm e^C$ is a nonzero constant.

We can see though that also when $C' = 0,$ we get a solution: $y(x) equiv 0.$ We therefore have a general solution $y(x) = C' e^x,$ where $C'$ is a constant.

share|cite|improve this answer

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

1

$begingroup$

Here comes a solution without "magical" manipulation of differentials.

Assuming that $y neq 0$ on some interval, divide the equation with $y$:
$$frac{y'(x)}{y(x)} = 1.$$

Now the left hand side can be written as the derivative of $ln |y(x)|$ and the right hand side as the derivative of $x$:
$$frac{d}{dx} ln |y(x)| = frac{d}{dx} x.$$

Therefore, $ln |y(x)| = x + C,$ where $C$ is some constant, and on the interval we thus have
$$y(x) = pm e^{x+C} = pm e^C e^x = C' e^x,$$
where $C' = pm e^C$ is a nonzero constant.

We can see though that also when $C' = 0,$ we get a solution: $y(x) equiv 0.$ We therefore have a general solution $y(x) = C' e^x,$ where $C'$ is a constant.

share|cite|improve this answer

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

$endgroup$

add a comment | 

$begingroup$

Here comes a solution without "magical" manipulation of differentials.

Assuming that $y neq 0$ on some interval, divide the equation with $y$:
$$frac{y'(x)}{y(x)} = 1.$$

Now the left hand side can be written as the derivative of $ln |y(x)|$ and the right hand side as the derivative of $x$:
$$frac{d}{dx} ln |y(x)| = frac{d}{dx} x.$$

Therefore, $ln |y(x)| = x + C,$ where $C$ is some constant, and on the interval we thus have
$$y(x) = pm e^{x+C} = pm e^C e^x = C' e^x,$$
where $C' = pm e^C$ is a nonzero constant.

We can see though that also when $C' = 0,$ we get a solution: $y(x) equiv 0.$ We therefore have a general solution $y(x) = C' e^x,$ where $C'$ is a constant.

share|cite|improve this answer

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

$endgroup$

share|cite|improve this answer

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

answered Mar 10 at 16:50

md2perpemd2perpe

8,19611028

1

$begingroup$

The first question in your title is

What are differential equations?

The informal answer is that often what we know about a quantity we are interested in is its rate of change. For example, if you know the velocity as a function $v(t)$ of time you can find the position $p$ by solving the differential equation
$$
p'(t) = v(t) .
$$
That's just integration. You use any known position to find the "constant of integration".

The differential equation you ask about,
$$
y' = y,
$$
says that the rate of change of $y$ is $y$ itself. In slightly more generality
$$
y' = ky
$$
says that the rate of change of $y$ is proportional to $y$. That classic equation models population growth over time or compound interest: the more stuff there is the faster it's growing. The solution is exponential growth:
$$
y = Ce^{kt}.
$$

The second order differential equation
$$
y" = -ky
$$
models a spring: the second derivative of position is (essentially) a force (that's Newton's law $F=ma$). The equation says that the farther $y$ is from $0$ the greater the force returning it to $0$. The solutions are the oscillating functions $y = sin(t+ phi)$.

share|cite|improve this answer

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

$endgroup$

add a comment | 

1

$begingroup$

The first question in your title is

What are differential equations?

The informal answer is that often what we know about a quantity we are interested in is its rate of change. For example, if you know the velocity as a function $v(t)$ of time you can find the position $p$ by solving the differential equation
$$
p'(t) = v(t) .
$$
That's just integration. You use any known position to find the "constant of integration".

The differential equation you ask about,
$$
y' = y,
$$
says that the rate of change of $y$ is $y$ itself. In slightly more generality
$$
y' = ky
$$
says that the rate of change of $y$ is proportional to $y$. That classic equation models population growth over time or compound interest: the more stuff there is the faster it's growing. The solution is exponential growth:
$$
y = Ce^{kt}.
$$

The second order differential equation
$$
y" = -ky
$$
models a spring: the second derivative of position is (essentially) a force (that's Newton's law $F=ma$). The equation says that the farther $y$ is from $0$ the greater the force returning it to $0$. The solutions are the oscillating functions $y = sin(t+ phi)$.

share|cite|improve this answer

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

$endgroup$

add a comment | 

$begingroup$

The first question in your title is

What are differential equations?

The informal answer is that often what we know about a quantity we are interested in is its rate of change. For example, if you know the velocity as a function $v(t)$ of time you can find the position $p$ by solving the differential equation
$$
p'(t) = v(t) .
$$
That's just integration. You use any known position to find the "constant of integration".

The differential equation you ask about,
$$
y' = y,
$$
says that the rate of change of $y$ is $y$ itself. In slightly more generality
$$
y' = ky
$$
says that the rate of change of $y$ is proportional to $y$. That classic equation models population growth over time or compound interest: the more stuff there is the faster it's growing. The solution is exponential growth:
$$
y = Ce^{kt}.
$$

The second order differential equation
$$
y" = -ky
$$
models a spring: the second derivative of position is (essentially) a force (that's Newton's law $F=ma$). The equation says that the farther $y$ is from $0$ the greater the force returning it to $0$. The solutions are the oscillating functions $y = sin(t+ phi)$.

share|cite|improve this answer

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

$endgroup$

share|cite|improve this answer

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120

answered Mar 11 at 1:28

Ethan BolkerEthan Bolker

45k553120