Using fsolve to solve non-linear equations

In this screencast, we go over a chemical engineering example of using Matlab's fsolve to solve for unknown composition, molar flow rates, and temperature in a system of non-linear equations associated with a condenser.
To achieve a higher selectivity of C with respect to D, you would like to have a smaller reactor. This is because over time, C will degrade into D (A + C goes to D), and in a larger reactor, the components spend a longer time in the reactor, allowing for more time for this degradation process to take place. A smaller reactor will have time to make C, but not as much time for that to change into D.

[@emlmm88]

Wow, it's uncanny. This is almost the exact problem that our ChemE professor gave us and was giving me so much trouble. Thank you!

[@annicafreytag1216]

Thank you so much! This is great. Helped me a lot!

[@priyajeni5560]

sir is it possible to solve the below type of equations using fsolve... if it is how to give an input in MATLAB? A1=(1+1/beta)D^2+diag(f+g)D-2h*I B1=2p^2 where beta=0.8 c=0.3 Initial guess is f(0)=1-e^(-n) g(0)=c*(1-e^-n) p(0)=e^(-n) h(0)=c*e^(-n) This is Spectral collocation method...i have no idea to give input as matrix form... pls kindly help..

[@drgregoryreeves]

Hi Priya, I am having a hard time determining what your question is. So it looks like you have two equations and four unknowns? I don't remember if fsolve can work with underdetermined systems. Do you have other constraints you can work with? To put your two equations into matrix form, the last line of your function m-file can be F = [A1; B1], and then the output of the function m-file should be "F". That function m-file will also expect a vector of your unknown variables as the first input (call it "Y"). So in the first few lines of the m-file, you will need to say something like: f = Y(1); g = Y(2); p = Y(3); h = Y(4);

[@HUNDANOL]

This is very helpful. Thanks! Can you suggest me a book that describes fsolve. I am using fsolve for much complicated system of equations, more than 200 equations and for different parameter values.

[@Perspective8847]

Thanks for the video. It's very close to what I'm looking for, i would like you to help me out. I'm working on alloys and need to calculate the value of a parameter that has 3 variables that are also equations, the parameter and one of the variables have 4 variables that i need to afterwards vary to get the values of the parameter over a range of concentration fitting experimental curve

[@olawalebello7593]

Hello, I wish you could solve similar problem with six unknown parameters using "lsqnonlin function"

[@doc_microscope]

Good afternoon, i have a problem: I have this linear system of 4 equations in four unknows P1, P2 P3 and P4: a1*P1+a2*P2+a3*P3+a4*P4=a; b1*P1+b2*P2+b3*P3+b4*P4=b; c1*P1+c2*P2+c3*P3+c4*P4=c; d1*P1+d2*P2+d3*P3+d4*P4=d; a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4,d1,d2,d3,d4 are constants and a,b,c,d are the known terms. How i can implement it on Matlab such that to find the roots?

[@doc_microscope]

It isn't linear equations. The equations are these: G1=10log10(P_out1/Pin1)=10log10exp( a1*P1+a2*P2+a3*P3+a4*P4-a) G2=10log10(P_out2/Pin2)=10log10exp(b1*P1+b2*P2+b3*P3+b4*P4-b) etc. Is it possible approximate the logarithm in these manner? 10log10(P_out/Pin)=10log(e)*( a1*P1+a2*P2+a3*P3+a4*P4-a) such that i can to obtain a linear system?

[@rishijaiswal1354]

Why is there a dot in 10^.(A-B/C+T)?

[@rishijaiswal1354]

Thankyou very much!

[@FcoJRueda]

What's the book?

[@FcoJRueda]

Where it is written all the exercises? Ahhhhh and where can I find that document?

[@fangirl8864]

Sir, how could we solve non linear equations of two variables let's say equations -x1+2x1^3+x2=0,-x1-x2=0

[@GregReeves1520]

sorry, the notification about your message went to my spam box, for some reason. I have to say I am not totally sure what you're asking. If you follow the fsolve example above, but input your two equations instead of the ones used in the example, you should be able to solve them. Alternatively, from looking at your equations, the second equation is x1 = -x2, so you should be able to substitute that relationship into the firs equation and get a cubic, which you can solve analytically or numerically very easily with Matlab's "roots" function.