How can we model this equation in simulink dx/dy + x + y = 1 where x is the input and y is the output.

how to model this equation in simulink dx/dy + x + y = 1

[John BG]

BACK TO BASICS

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here I am using the grey background to highlight functions, not MATLAB code.

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Misha and Basa,

the usage of 1/u in Misha's solution, in this context is erroneous, let me show you why:

1.- one of the possible solutions to dx/dy+y+x-1=0 is

x=1-exp(-y)

2.- inverting x(y)

x=1-exp(-y) 

ln(1-x)=-y

so we have

y=-ln(1-x), or ln(|1-x|)

the 1st key point you need to understand is that d(y)/dx is not a division operator, it's a convention to express derivative, which is an operator that generally speaking does not comply with

if u=f(v) d(f(v),v)=k 
let be v=g(u) 
it's not generally correct assume d(g(u),u)=k/f(v)

2.- according to Misha,

dy/dx=1/(1-x-y)

but

dy/dx=d(-ln(1-x))/dx=1/(1-x)

yet

1/(1-x-y)=1/(1-x+ln(1-x)) or 1/(1-ln(|1-x|))

which obviously are not the same

Misha the 2nd reason why you shouldn't be using 1/u is that you may be losing poles while constraining the function domain to that of u=f(v), not the variable intended to be the domain.

For these reasons I consider my answer deserves to be the accepted answer.

Additionally, Basa, I gave the Simulink diagram first, Misha had just given some general indications, Misha saw my diagram, he builds an erroneous Simulink equation, and you Basa mark as accepted an erroneous Simulink circuit for the above reasons.

Please reconsider marking my answer as accepted.

John BG

<mailto:[e-mail] [e-mail]>

[Roger Stafford]

If it is of interest to you, there is a general solution to this differential equation which expresses x as a function of y:

   x = 2 - y + K*exp(-y)

where K is an arbitrary constant of integration. Expressing y as a function of x would undoubtedly involve the ‘lambertw’ function.

[John BG]

You are trying to get off the tangent with a technical term, let's do it:

1.-

x = 2 - y + K*exp(-y)

dx/dy=-1-K*exp(-y)

dx/dy+x+y-1=-1-K*exp(-y)   +2 - y + K*exp(-y)  -1 +y = 0   correct

now

x-2 = -y+K*exp(-y)

x2=x-2

x2 is a particular case of

x3=p^y-a*y

replace x with y and b with x in Example 1

https://en.wikipedia.org/wiki/Lambert_W_function

line starting with 'More generally, ..' p would be number e, 2.7182.. at least you will agree upon this, won't you?

substitution:

-t=y+x3/a

further substitution:

R=-1/a*p^(x/a)

then

t=W(R*ln(p))/ln(p)

because p is e, then

t=W(R)

y=-W(-1/a*exp(-x/a))-x/a

now back on track with your W technicality

dy/dx=-1/a-d(W(-1/a*exp(-x/a)))/dx

dW(z)/dz=W/(z*(1+W)) for z~=-1/e

dy/dx=-1/a-WW(-1/a*exp(-x/a))/(W(-1/a*exp(-x/a))*(1+W(-1/a*exp(-x/a))))

so? where is simple inversion of the operator / ??

Now

1.- fix your answer

2.- go to Basa and tell the kid

and don't tell not to quote from Wikipedia, it's where you got your 'lambertw' from, isn't it?

[Raghad]

It is required to solve the following dynamic equation on Matlab/Simulink. submit a report in word for what you have done and submit the Simulink file.

3 dy/dt +2y = x(t)

x(t)=1, y(0)=1

Answer 1

[Mischa Kim]

Re-write as

 dx/dy = 1 - x - y  % y is your independent variable, sort of "t"

therefore

 dy/dx = 1/(1 - x - y)

Then, from left to right (in your Simulink block diagram) you have

• a summation block, adding up the right hand side, and taking the reciprocal value
• the output goes into an integrator block that outputs y, which is the solution and also feeds back into the summation block

[Basavalingappa Mudhol]

i should have y as the output from my system and x being my input i think this is the other case that you have put up in simulink isnt it? Please clarify..

[Mischa Kim]

Right you are. I updated the diagram, see above.

[John BG]

BACK TO BASICS

.

here I am using the grey background to highlight functions, not MATLAB code.

.

Misha and Basa,

the usage of 1/u in Misha's solution, in this context is erroneous, let me show you why:

1.- one of the possible solutions to dx/dy+y+x-1=0 is

x=1-exp(-y)

2.- inverting x(y)

x=1-exp(-y) 

ln(1-x)=-y

so we have

y=-ln(1-x), or ln(|1-x|)

the 1st key point you need to understand is that d(y)/dx is not a division operator, it's a convention to express derivative, which is an operator that generally speaking does not comply with

if u=f(v) d(f(v),v)=k 
let be v=g(u) 
it's not generally correct assume d(g(u),u)=k/f(v)

2.- according to Misha,

dy/dx=1/(1-x-y)

but

dy/dx=d(-ln(1-x))/dx=1/(1-x)

yet

1/(1-x-y)=1/(1-x+ln(1-x)) or 1/(1-ln(|1-x|))

which obviously are not the same

Misha the 2nd reason why you shouldn't be using 1/u is that you may be losing poles while constraining the function domain to that of u=f(v), not the variable intended to be the domain.

For these reasons I consider my answer deserves to be the accepted answer.

John BG

<mailto:[e-mail] [e-mail]>

Answer 2

[John BG]

Hi Basa

got you this

the upload does not accept Simulink file name extensions (!?) people compress and upload but it should be straight forward to upload MATLAB related files with MATLAB file name extensions, shouldn't it?

Well, if this answer helps you with this question, would you please be so kind to mark my answer as accepted?

thanks for attention, awaiting answer

John BG <mailto:[e-mail] [e-mail]>

[Mischa Kim]

Usage of the derivative block is not a recommended best practice unless absolutely necessary. For more information, check out this answer.

[John BG]

Misha please have a look to my comments, the usage of 1/u to invert a derivative is completely erroneous as a I show above. You just found something that algebraically looked like it would work, you plugged it an got an erroneous Simulink circuit, that Basa took as good answer.

Please review you answer and correct accordingly.