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0

$begingroup$

I'm a beginner in using Matlab. if I have a DE like

$$x'(t)=frac1sin (2x)$$

How I can Implement in Matlab to calculate just answer?

I try dsolve like as:

ySol(t) = dsolve(ode,cond)
but couldn't define ode, and cond.

Update: this is my matlab:

and different answer from Wolfram:

ordinary-differential-equations matlab wolfram-alpha

share|cite|improve this question

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

$endgroup$

add a comment | 

0

$begingroup$

I'm a beginner in using Matlab. if I have a DE like

$$x'(t)=frac1sin (2x)$$

How I can Implement in Matlab to calculate just answer?

I try dsolve like as:

ySol(t) = dsolve(ode,cond)
but couldn't define ode, and cond.

Update: this is my matlab:

and different answer from Wolfram:

ordinary-differential-equations matlab wolfram-alpha

share|cite|improve this question

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

$endgroup$

add a comment | 

$begingroup$

I'm a beginner in using Matlab. if I have a DE like

$$x'(t)=frac1sin (2x)$$

How I can Implement in Matlab to calculate just answer?

I try dsolve like as:

ySol(t) = dsolve(ode,cond)
but couldn't define ode, and cond.

Update: this is my matlab:

and different answer from Wolfram:

ordinary-differential-equations matlab wolfram-alpha

share|cite|improve this question

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

$endgroup$

share|cite|improve this question

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

share|cite|improve this question

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

edited Mar 7 at 8:36

Dylan

13.7k31027

edited Mar 7 at 8:36

Dylan

13.7k31027

asked Mar 6 at 23:05

user355834user355834

416

asked Mar 6 at 23:05

user355834user355834

416

2 Answers
2

active

oldest

votes

2

$begingroup$

These are the same answer. Separate to get

$$ sin(2x) dx = dt $$

There are two ways to do this. The first is to integrate directly:

$$ -frac12cos(2x) = t + c_1 $$
$$ cos(2x) = -2(t+c_1) $$
$$ x = frac12 arccos(-2(t+c_1)) $$

This is the answer given by Wolfram.

The second way is to use the double-angle formula

$$ 2sin(x)cos(x) dx = dt $$

$$ sin^2(x) = t + c_2 $$

$$ sin (x) = sqrtt+c_2 $$

$$ x = arcsin(sqrtt+c_2) $$

This is the answer given by MATLAB.

If you're wondering why there two different anti-derivatives of $sin(2x)$, it's because they differ by a constant

$$ sin^2 x = -frac12cos (2x) + frac12 $$

share|cite|improve this answer

edited Mar 7 at 9:00

answered Mar 7 at 8:41

DylanDylan

13.7k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [+1] Very didactic answer. Two points are still puzzling for me 1) On the asker side : an initial condition $x(0)=1$ has been introduced. Why ? 2) On the answerer side : why don't you consider as well $sin(x)=-sqrtt+c_2$ ?
  $endgroup$
  – Jean Marie
  Mar 7 at 8:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie 1) MATLAB requires an initial condition to solve the problem, so I assumed $x(0)=1$ was put in there as a placeholder. 2) I didn't feel like going into details, since the point was just to illustrate that the 2 answers are more or less equivalent. If you want to be pedantic, the sign of the square root depends on the initial condition given.
  $endgroup$
  – Dylan
  Mar 7 at 8:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you use the adjective "pedantic" ? Conventionally, working with real numbers, symbol $sqrt...$ designates a positive quantity, and $-sqrt...$ is for a negative quantity which is different.
  $endgroup$
  – Jean Marie
  Mar 7 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie You take the positive $sqrtcdots$ if the initial condition is positive and the negative $-sqrtcdots$ if the initial condition is negative.
  $endgroup$
  – Dylan
  Mar 7 at 9:18
  
  

  
  
  
  

add a comment | 

1

$begingroup$

A (Matlab oriented) complement to the very didactic answer by @Dylan.

Here is an extension of your program displaying either two or one solution(s). I use some Matlab instructions that may be new to you, in particular with a symbolic initial value $a$ (you had fixed this initial value to $a=1$) that can be constrained to be positive. The first part is with symbolic variables, almost the same as yours ; the second part uses numerical variables in order to see what happens for different values of $a$:

clear all;close all;
syms x(t) a
ode=diff(x)==1/sin(2*x); % no need to take diff(x,t)...
cond=x(0)==a; % a instead of 1
%assume(a>0)
s=dsolve(ode,cond) : % one or two solutions
%%%
T=0:0.01:0.7;
for a=0:0.1:pi/2
 g=inline(s(1));plot(T,real(g(a,T)),'r');hold on;
 g=inline(s(2));plot(T,real(g(a,T)),'b');hold on;% to be suppressed if a>0
end;

The answer(s) given by Matlab are :

s =

asin((sin(a)^2 + t)^(1/2))

-asin((sin(a)^2 + t)^(1/2))

Which one is the good one ?... Both...

If we introduce "assume(a>0)", Matlab gives a single answer which is the first one.

Other analytical facts could be commented (for example the domains of existence of solutions), but I think the asker will be satisfied with the details already given.

share|cite|improve this answer

edited 2 days ago

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

$endgroup$

add a comment | 

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2

$begingroup$

These are the same answer. Separate to get

$$ sin(2x) dx = dt $$

There are two ways to do this. The first is to integrate directly:

$$ -frac12cos(2x) = t + c_1 $$
$$ cos(2x) = -2(t+c_1) $$
$$ x = frac12 arccos(-2(t+c_1)) $$

This is the answer given by Wolfram.

The second way is to use the double-angle formula

$$ 2sin(x)cos(x) dx = dt $$

$$ sin^2(x) = t + c_2 $$

$$ sin (x) = sqrtt+c_2 $$

$$ x = arcsin(sqrtt+c_2) $$

This is the answer given by MATLAB.

If you're wondering why there two different anti-derivatives of $sin(2x)$, it's because they differ by a constant

$$ sin^2 x = -frac12cos (2x) + frac12 $$

share|cite|improve this answer

edited Mar 7 at 9:00

answered Mar 7 at 8:41

DylanDylan

13.7k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [+1] Very didactic answer. Two points are still puzzling for me 1) On the asker side : an initial condition $x(0)=1$ has been introduced. Why ? 2) On the answerer side : why don't you consider as well $sin(x)=-sqrtt+c_2$ ?
  $endgroup$
  – Jean Marie
  Mar 7 at 8:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie 1) MATLAB requires an initial condition to solve the problem, so I assumed $x(0)=1$ was put in there as a placeholder. 2) I didn't feel like going into details, since the point was just to illustrate that the 2 answers are more or less equivalent. If you want to be pedantic, the sign of the square root depends on the initial condition given.
  $endgroup$
  – Dylan
  Mar 7 at 8:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you use the adjective "pedantic" ? Conventionally, working with real numbers, symbol $sqrt...$ designates a positive quantity, and $-sqrt...$ is for a negative quantity which is different.
  $endgroup$
  – Jean Marie
  Mar 7 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie You take the positive $sqrtcdots$ if the initial condition is positive and the negative $-sqrtcdots$ if the initial condition is negative.
  $endgroup$
  – Dylan
  Mar 7 at 9:18
  
  

  
  
  
  

add a comment | 

2

$begingroup$

These are the same answer. Separate to get

$$ sin(2x) dx = dt $$

There are two ways to do this. The first is to integrate directly:

$$ -frac12cos(2x) = t + c_1 $$
$$ cos(2x) = -2(t+c_1) $$
$$ x = frac12 arccos(-2(t+c_1)) $$

This is the answer given by Wolfram.

The second way is to use the double-angle formula

$$ 2sin(x)cos(x) dx = dt $$

$$ sin^2(x) = t + c_2 $$

$$ sin (x) = sqrtt+c_2 $$

$$ x = arcsin(sqrtt+c_2) $$

This is the answer given by MATLAB.

If you're wondering why there two different anti-derivatives of $sin(2x)$, it's because they differ by a constant

$$ sin^2 x = -frac12cos (2x) + frac12 $$

share|cite|improve this answer

edited Mar 7 at 9:00

answered Mar 7 at 8:41

DylanDylan

13.7k31027

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  [+1] Very didactic answer. Two points are still puzzling for me 1) On the asker side : an initial condition $x(0)=1$ has been introduced. Why ? 2) On the answerer side : why don't you consider as well $sin(x)=-sqrtt+c_2$ ?
  $endgroup$
  – Jean Marie
  Mar 7 at 8:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie 1) MATLAB requires an initial condition to solve the problem, so I assumed $x(0)=1$ was put in there as a placeholder. 2) I didn't feel like going into details, since the point was just to illustrate that the 2 answers are more or less equivalent. If you want to be pedantic, the sign of the square root depends on the initial condition given.
  $endgroup$
  – Dylan
  Mar 7 at 8:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you use the adjective "pedantic" ? Conventionally, working with real numbers, symbol $sqrt...$ designates a positive quantity, and $-sqrt...$ is for a negative quantity which is different.
  $endgroup$
  – Jean Marie
  Mar 7 at 9:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JeanMarie You take the positive $sqrtcdots$ if the initial condition is positive and the negative $-sqrtcdots$ if the initial condition is negative.
  $endgroup$
  – Dylan
  Mar 7 at 9:18
  
  

  
  
  
  

add a comment | 

$begingroup$

These are the same answer. Separate to get

$$ sin(2x) dx = dt $$

There are two ways to do this. The first is to integrate directly:

$$ -frac12cos(2x) = t + c_1 $$
$$ cos(2x) = -2(t+c_1) $$
$$ x = frac12 arccos(-2(t+c_1)) $$

This is the answer given by Wolfram.

The second way is to use the double-angle formula

$$ 2sin(x)cos(x) dx = dt $$

$$ sin^2(x) = t + c_2 $$

$$ sin (x) = sqrtt+c_2 $$

$$ x = arcsin(sqrtt+c_2) $$

This is the answer given by MATLAB.

If you're wondering why there two different anti-derivatives of $sin(2x)$, it's because they differ by a constant

$$ sin^2 x = -frac12cos (2x) + frac12 $$

share|cite|improve this answer

edited Mar 7 at 9:00

answered Mar 7 at 8:41

DylanDylan

13.7k31027

$endgroup$

share|cite|improve this answer

edited Mar 7 at 9:00

answered Mar 7 at 8:41

DylanDylan

13.7k31027

answered Mar 7 at 8:41

DylanDylan

13.7k31027

answered Mar 7 at 8:41

DylanDylan

13.7k31027

1

$begingroup$

A (Matlab oriented) complement to the very didactic answer by @Dylan.

Here is an extension of your program displaying either two or one solution(s). I use some Matlab instructions that may be new to you, in particular with a symbolic initial value $a$ (you had fixed this initial value to $a=1$) that can be constrained to be positive. The first part is with symbolic variables, almost the same as yours ; the second part uses numerical variables in order to see what happens for different values of $a$:

clear all;close all;
syms x(t) a
ode=diff(x)==1/sin(2*x); % no need to take diff(x,t)...
cond=x(0)==a; % a instead of 1
%assume(a>0)
s=dsolve(ode,cond) : % one or two solutions
%%%
T=0:0.01:0.7;
for a=0:0.1:pi/2
 g=inline(s(1));plot(T,real(g(a,T)),'r');hold on;
 g=inline(s(2));plot(T,real(g(a,T)),'b');hold on;% to be suppressed if a>0
end;

The answer(s) given by Matlab are :

s =

asin((sin(a)^2 + t)^(1/2))

-asin((sin(a)^2 + t)^(1/2))

Which one is the good one ?... Both...

If we introduce "assume(a>0)", Matlab gives a single answer which is the first one.

Other analytical facts could be commented (for example the domains of existence of solutions), but I think the asker will be satisfied with the details already given.

share|cite|improve this answer

edited 2 days ago

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

$endgroup$

add a comment | 

1

$begingroup$

A (Matlab oriented) complement to the very didactic answer by @Dylan.

Here is an extension of your program displaying either two or one solution(s). I use some Matlab instructions that may be new to you, in particular with a symbolic initial value $a$ (you had fixed this initial value to $a=1$) that can be constrained to be positive. The first part is with symbolic variables, almost the same as yours ; the second part uses numerical variables in order to see what happens for different values of $a$:

clear all;close all;
syms x(t) a
ode=diff(x)==1/sin(2*x); % no need to take diff(x,t)...
cond=x(0)==a; % a instead of 1
%assume(a>0)
s=dsolve(ode,cond) : % one or two solutions
%%%
T=0:0.01:0.7;
for a=0:0.1:pi/2
 g=inline(s(1));plot(T,real(g(a,T)),'r');hold on;
 g=inline(s(2));plot(T,real(g(a,T)),'b');hold on;% to be suppressed if a>0
end;

The answer(s) given by Matlab are :

s =

asin((sin(a)^2 + t)^(1/2))

-asin((sin(a)^2 + t)^(1/2))

Which one is the good one ?... Both...

If we introduce "assume(a>0)", Matlab gives a single answer which is the first one.

Other analytical facts could be commented (for example the domains of existence of solutions), but I think the asker will be satisfied with the details already given.

share|cite|improve this answer

edited 2 days ago

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

$endgroup$

add a comment | 

$begingroup$

A (Matlab oriented) complement to the very didactic answer by @Dylan.

Here is an extension of your program displaying either two or one solution(s). I use some Matlab instructions that may be new to you, in particular with a symbolic initial value $a$ (you had fixed this initial value to $a=1$) that can be constrained to be positive. The first part is with symbolic variables, almost the same as yours ; the second part uses numerical variables in order to see what happens for different values of $a$:

clear all;close all;
syms x(t) a
ode=diff(x)==1/sin(2*x); % no need to take diff(x,t)...
cond=x(0)==a; % a instead of 1
%assume(a>0)
s=dsolve(ode,cond) : % one or two solutions
%%%
T=0:0.01:0.7;
for a=0:0.1:pi/2
 g=inline(s(1));plot(T,real(g(a,T)),'r');hold on;
 g=inline(s(2));plot(T,real(g(a,T)),'b');hold on;% to be suppressed if a>0
end;

The answer(s) given by Matlab are :

s =

asin((sin(a)^2 + t)^(1/2))

-asin((sin(a)^2 + t)^(1/2))

Which one is the good one ?... Both...

If we introduce "assume(a>0)", Matlab gives a single answer which is the first one.

Other analytical facts could be commented (for example the domains of existence of solutions), but I think the asker will be satisfied with the details already given.

share|cite|improve this answer

edited 2 days ago

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

$endgroup$

clear all;close all;
syms x(t) a
ode=diff(x)==1/sin(2*x); % no need to take diff(x,t)...
cond=x(0)==a; % a instead of 1
%assume(a>0)
s=dsolve(ode,cond) : % one or two solutions
%%%
T=0:0.01:0.7;
for a=0:0.1:pi/2
 g=inline(s(1));plot(T,real(g(a,T)),'r');hold on;
 g=inline(s(2));plot(T,real(g(a,T)),'b');hold on;% to be suppressed if a>0
end;

share|cite|improve this answer

edited 2 days ago

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154

answered Mar 7 at 10:19

Jean MarieJean Marie

30.6k42154