Answered

Solving Multiple First Order Differential Equations for Constants

Replace the time derivatives by difference quotients. Then you get a linear system of equations A*[k1;k2;k3;k4;k5;k6] = b wh...

Solving Multiple First Order Differential Equations for Constants

Replace the time derivatives by difference quotients. Then you get a linear system of equations A*[k1;k2;k3;k4;k5;k6] = b wh...

2 dagen ago | 0

Answered

Evaluating a complex integral

format long fun = @(t,x,y) exp(-t.^4 + 1i.*y.*t - x.*t.^2 + 1i*pi*0.125); P = @(x,y) integral(@(t)fun(t,x*exp(-1i*pi*0.25)...

Evaluating a complex integral

format long fun = @(t,x,y) exp(-t.^4 + 1i.*y.*t - x.*t.^2 + 1i*pi*0.125); P = @(x,y) integral(@(t)fun(t,x*exp(-1i*pi*0.25)...

2 dagen ago | 0

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Answered

Is it possibile to use fsolve(fun,x0) giving input to fun?

fun = @(x)root2d(x,a,b,c,d,e,f)

Is it possibile to use fsolve(fun,x0) giving input to fun?

fun = @(x)root2d(x,a,b,c,d,e,f)

2 dagen ago | 0

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Answered

Summation with FOR loop question

format long fak = 1.0; s = fak; for i = 1:1600 fak = fak/i; s = s + fak; end s exp(1)

Summation with FOR loop question

format long fak = 1.0; s = fak; for i = 1:1600 fak = fak/i; s = s + fak; end s exp(1)

3 dagen ago | 1

Answered

How to make polynom with unknown coefficients?

n = 10; c = sym('c', [1 n]); p = poly2sym(c)

How to make polynom with unknown coefficients?

n = 10; c = sym('c', [1 n]); p = poly2sym(c)

3 dagen ago | 0

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Answered

How to solve a complex equation which has one equation but two real variables?

Does that help ? >> syms z k >> z0 = 1+3*1I; >> a = solve(z==k*z0,z) a = (sym) k*(1 + 3*I) >> real(a) ans = (sym) re(k) - ...

How to solve a complex equation which has one equation but two real variables?

Does that help ? >> syms z k >> z0 = 1+3*1I; >> a = solve(z==k*z0,z) a = (sym) k*(1 + 3*I) >> real(a) ans = (sym) re(k) - ...

3 dagen ago | 0

Answered

Double or higher order integration by numerical method

Divide the y-interval of integration into subintervals ystart = y1 < y2 < y3 < ... < yn = yend and call your function "traorl"...

Double or higher order integration by numerical method

Divide the y-interval of integration into subintervals ystart = y1 < y2 < y3 < ... < yn = yend and call your function "traorl"...

4 dagen ago | 1

Answered

Symsum and sum output different result

Try n = 1:10000; lamda = n*pi/a; pih = ((lamda+h/k1)-(lamda-h/k1).*exp(-2*lamda*t1))./((lamda+h/k1)+(lamda-h/k1).*exp(-2*lamd...

Symsum and sum output different result

Try n = 1:10000; lamda = n*pi/a; pih = ((lamda+h/k1)-(lamda-h/k1).*exp(-2*lamda*t1))./((lamda+h/k1)+(lamda-h/k1).*exp(-2*lamd...

5 dagen ago | 0

Answered

How to iterate until convergence?

The structure of your code should look like: function main V = linspace(0.01,0.25,1000); R = zeros(numel(V),1); R0 =...

How to iterate until convergence?

The structure of your code should look like: function main V = linspace(0.01,0.25,1000); R = zeros(numel(V),1); R0 =...

5 dagen ago | 0

Answered

How to add limit condition to existing code

If you are not satisfied with the approximation for h=0.01 in the code, you could evaluate the function fgl_deriv for two or thr...

How to add limit condition to existing code

If you are not satisfied with the approximation for h=0.01 in the code, you could evaluate the function fgl_deriv for two or thr...

5 dagen ago | 0

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Answered

How to solve a Partial Differential Ecuation?

The code at the end of the link solves a single equation. MATLAB does not have the ability to symbolically solve PDEs. If you w...

How to solve a Partial Differential Ecuation?

The code at the end of the link solves a single equation. MATLAB does not have the ability to symbolically solve PDEs. If you w...

6 dagen ago | 0

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Answered

Plotting implicit equation with fimplicit

function main fimplicit (@(x,y)f(x,y)) end function fun = f(x,y) n = 8; a1 = 1.0086*y - 0.9216*(x - y); b1 = 1.0107*(...

Plotting implicit equation with fimplicit

function main fimplicit (@(x,y)f(x,y)) end function fun = f(x,y) n = 8; a1 = 1.0086*y - 0.9216*(x - y); b1 = 1.0107*(...

6 dagen ago | 0

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Answered

What are the boundary conditions of interp1 function?

MATLAB uses a "not-a-knot" end condition for splines.

What are the boundary conditions of interp1 function?

MATLAB uses a "not-a-knot" end condition for splines.

6 dagen ago | 1

Answered

How can numerically compute eigenvalues of an ordinary differential equation in MATLAB?

https://www.wolframalpha.com/input/?i=y''(x)%2B1%2Fx*y'(x)%2Ba*(1-x%5E2)*y(x)%3D0,+y'(0)%3D0 So you are left with the problem t...

How can numerically compute eigenvalues of an ordinary differential equation in MATLAB?

https://www.wolframalpha.com/input/?i=y''(x)%2B1%2Fx*y'(x)%2Ba*(1-x%5E2)*y(x)%3D0,+y'(0)%3D0 So you are left with the problem t...

6 dagen ago | 0

Answered

How to solve a Partial Differential Ecuation?

Here is code to start from: https://de.mathworks.com/matlabcentral/answers/440015-first-order-partial-differential-equations-sy...

How to solve a Partial Differential Ecuation?

Here is code to start from: https://de.mathworks.com/matlabcentral/answers/440015-first-order-partial-differential-equations-sy...

6 dagen ago | 0

Answered

Calculate derivative of erfcx(x) function

Try x = 0:0.1:5; analytical_derivative = -2/sqrt(pi) + 2.*x.*erfcx(x); numerical_derivative = (erfcx(x+1.0e-6)-erfcx(x))*1e6;...

Calculate derivative of erfcx(x) function

Try x = 0:0.1:5; analytical_derivative = -2/sqrt(pi) + 2.*x.*erfcx(x); numerical_derivative = (erfcx(x+1.0e-6)-erfcx(x))*1e6;...

9 dagen ago | 0

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Answered

Taking the second derivative

second_der = zeros(size(V_preamp)); second_der(2:end-1) = ((V_preamp(3:end)-V_preamp(2:end-1))./... (M(...

Taking the second derivative

second_der = zeros(size(V_preamp)); second_der(2:end-1) = ((V_preamp(3:end)-V_preamp(2:end-1))./... (M(...

10 dagen ago | 1

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Answered

Beam Equation - problem with boundary condition

v = vx = 0 at x=L w = wx = 0 at x=0 ?

Beam Equation - problem with boundary condition

v = vx = 0 at x=L w = wx = 0 at x=0 ?

10 dagen ago | 0

Answered

Extrapolate best-fit line

If you have the coefficients c(1) and c(2) of the fit line l(x)=c(1)+c(2)*x, just evaluate the equation at the x-coordinate of t...

Extrapolate best-fit line

If you have the coefficients c(1) and c(2) of the fit line l(x)=c(1)+c(2)*x, just evaluate the equation at the x-coordinate of t...

10 dagen ago | 0

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Answered

How to plot a function of two variable

https://de.mathworks.com/help/matlab/ref/surf.html

How to plot a function of two variable

https://de.mathworks.com/help/matlab/ref/surf.html

10 dagen ago | 0

Answered

solving non linear differential equations

a = -0.2; b = 0.1; c = 0.1; epsilon = 1.0; fun = @(t,y)(t*y*(y-a)*(y-b)+c)/epsilon; tspan = [0 3.5]; y0 = 0; [T,Y] = ode4...

solving non linear differential equations

a = -0.2; b = 0.1; c = 0.1; epsilon = 1.0; fun = @(t,y)(t*y*(y-a)*(y-b)+c)/epsilon; tspan = [0 3.5]; y0 = 0; [T,Y] = ode4...

10 dagen ago | 0

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Answered

How to find the differential equation ? ode solver

If the output plot function is f(t) and f(t0)=f0, then the corresponding ODE reads dy/dt = f'(t) with y(t0) = f0. This ODE sh...

How to find the differential equation ? ode solver

If the output plot function is f(t) and f(t0)=f0, then the corresponding ODE reads dy/dt = f'(t) with y(t0) = f0. This ODE sh...

12 dagen ago | 1

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Answered

ODE45 error message "Error using exist The first input to exist must be a string scalar or character vector"

param = fminsearch(@lau,param0) [t,w] = ode45(@grinding,[t0 tp],w0)

ODE45 error message "Error using exist The first input to exist must be a string scalar or character vector"

param = fminsearch(@lau,param0) [t,w] = ode45(@grinding,[t0 tp],w0)

12 dagen ago | 1

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Answered

Fsolve for a generalizable number of equations.

A = [1, 0, 0, 0, 1/sub(1,1), 1/sub(2,1), 0, 0;... 0, 1, 0, 0, 1/sub(1,2), 1/sub(2,2), 1/sub(3,2), 0;... 0, 0, 1, 0, ...

Fsolve for a generalizable number of equations.

A = [1, 0, 0, 0, 1/sub(1,1), 1/sub(2,1), 0, 0;... 0, 1, 0, 0, 1/sub(1,2), 1/sub(2,2), 1/sub(3,2), 0;... 0, 0, 1, 0, ...

12 dagen ago | 0

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Answered

two-step Adams Moulton method

In the Adams-Moulton formula, y(i) appears on both sides of the equation. This means that the Adams-Moulton method is implicit. ...

two-step Adams Moulton method

In the Adams-Moulton formula, y(i) appears on both sides of the equation. This means that the Adams-Moulton method is implicit. ...

12 dagen ago | 0

Answered

Numerical integration twice wrt same variable

fun1 = @(x)integral(@(r)((r+1).^2+1)./r,1,x); fun2 = integral(@(x)(1+fun1(x)),0,1,'ArrayValued',true)

Numerical integration twice wrt same variable

fun1 = @(x)integral(@(r)((r+1).^2+1)./r,1,x); fun2 = integral(@(x)(1+fun1(x)),0,1,'ArrayValued',true)

12 dagen ago | 0

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Answered

Uniform distribution command for specific mean and standard deviation

(mean-sqrt(12)/2*s.d.) + sqrt(12)*s.d.*rand(N,1) is uniformly distributed on [a:b] = [m-sqrt(12)/2*s.d.:m+sqrt(12)/2*s.d.] ...

Uniform distribution command for specific mean and standard deviation

(mean-sqrt(12)/2*s.d.) + sqrt(12)*s.d.*rand(N,1) is uniformly distributed on [a:b] = [m-sqrt(12)/2*s.d.:m+sqrt(12)/2*s.d.] ...

12 dagen ago | 1

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Answered

Error calculating matrix operations

for i=1:14 ACCANG(:,i) = (-(Jinv))*([AB*(w2(i))*cosd(beta(i)), BO*(w3(i))*cosd(gamma(i)); -AB*(w2(i))*sind(beta(i)), -BO*(w3(i)...

Error calculating matrix operations

for i=1:14 ACCANG(:,i) = (-(Jinv))*([AB*(w2(i))*cosd(beta(i)), BO*(w3(i))*cosd(gamma(i)); -AB*(w2(i))*sind(beta(i)), -BO*(w3(i)...

13 dagen ago | 1

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Answered

How can I plot ode 45 for different values of a constant after specific time?

function main T = 1:1:3100; del = 0.0; [t1,y1] = ode45(@(t,y)f_1(t,y,del),T,C); T = 3100:1:3500; del = 1.0; C ...

How can I plot ode 45 for different values of a constant after specific time?

function main T = 1:1:3100; del = 0.0; [t1,y1] = ode45(@(t,y)f_1(t,y,del),T,C); T = 3100:1:3500; del = 1.0; C ...

13 dagen ago | 1

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Answered

Absolute Value in optimization expression

min: eps + (a*x + b*y + c*z - d) - eps <= 0 - (a*x + b*y + c*z - d) - eps <= 0 subject to x,y,z are integers in some range wi...

Absolute Value in optimization expression

min: eps + (a*x + b*y + c*z - d) - eps <= 0 - (a*x + b*y + c*z - d) - eps <= 0 subject to x,y,z are integers in some range wi...

13 dagen ago | 1