# How do I solve a second order non linear differential equation using matlab.

23 visualizzazioni (ultimi 30 giorni)
Patrick Guarente il 25 Set 2017
Risposto: Lewis Fer il 10 Giu 2021
I have a fluid dynamics problem and I need to derive an equation for motion.
After applying Newtons second law to the system, and replaceing all the constants with A and B. My equation looks like this.
z'' + A(z')^2 = B
With A and B both being constants.
Initial conditions being that z(0)=0, and z'(0)=0
And I need to solve for z(t).
Thank you
##### 4 CommentiMostra 2 commenti meno recentiNascondi 2 commenti meno recenti
James Tursa il 25 Set 2017
Modificato: James Tursa il 25 Set 2017
"... I need to find the equation for all time ..."
Are you looking for an analytical/symbolic solution? I thought that you simply wanted a numerical solution given your initial starting values.
Patrick Guarente il 25 Set 2017
Yes the analytical solution in terms of A and B.

Accedi per commentare.

### Risposte (4)

Teja Muppirala il 26 Set 2017
syms z(t) t A B
zp = diff(z,t);
zpp = diff(z,t,2);
eqn = ( zpp + A*zp^2 == B );
cond = [z(0)==0, zp(0)==0];
zSol = dsolve(eqn,cond,'IgnoreAnalyticConstraints',true);
zSol = unique(simplify(zSol));
This gives 3 solutions:
zSol =
log((C15*sinh(A^(1/2)*B^(1/2)*(t + A*B^(1/2)*1i)))/B^(1/2))/A
log(-(C18*sinh(A^(3/2)*B*1i - A^(1/2)*B^(1/2)*t))/B^(1/2))/A
log(cosh(A^(1/2)*B^(1/2)*t))/A
The first two look weird, but are valid solutions involving complex-valued z. The 3rd solution is real, and that's probably the one that you are looking for.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Accedi per commentare.

Lewis Fer il 10 Giu 2021
Hello, I am having troubles solving a system of second order nonlinear equations with boundary conditions using MATALB
Here is the equations:
f''(t)=3*f(t)*g(t) -g(t)+5*t;
g''(t)=-4f(t)*g(t)+f(t)-7*t;
the boundary conditions are: f'(0)=0 et h'(o)=5;
g(0)=3 et h'(2)=h(2)
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Accedi per commentare.

James Tursa il 25 Set 2017
Modificato: James Tursa il 25 Set 2017
Define a 2-element vector y:
y(1) = z
y(2) = z'
then solve your 2nd order ODE for the highest derivative:
z'' + A(z')^2 = B ==>
z'' = - A(z')^2 + B
then calculate the y element derivative equations, using this z derivative info:
d y(1) = d z = z' = y(2)
d y(2) = d z' = z'' = -A(z')^2 + B = -A*y(2) + B
So create a derivative function based on those two equations, using the function signature that you will find in the ode45 doc. Then call it using the outline provided in the example in the doc.
EDIT: SYMBOLIC SOLUTION
>> dsolve('D2z + A*(Dz)^2 = B')
ans =
C29 + (B^(1/2)*t)/A^(1/2)
C27 - (B^(1/2)*t)/A^(1/2)
log((exp(2*A^(3/2)*B^(1/2)*(C24 + t/A)) - 1)/(2*B^(1/2)*exp(A*(C16 + A^(1/2)*B^(1/2)*(C24 + t/A)))))/A
log((exp(2*A^(3/2)*B^(1/2)*(C20 - t/A)) - 1)/(2*B^(1/2)*exp(A*(C16 + A^(1/2)*B^(1/2)*(C20 - t/A)))))/A
##### 1 CommentoMostra -1 commenti meno recentiNascondi -1 commenti meno recenti
Patrick Guarente il 25 Set 2017
I'm not sure where those two equations will be implemented.

Accedi per commentare.

Torsten il 26 Set 2017
According to MATHEMATICA, the analytical solution is
z(x) = log(cosh(sqrt(A*B)*x))/A
Best wishes
Torsten.
##### 0 CommentiMostra -2 commenti meno recentiNascondi -2 commenti meno recenti

Accedi per commentare.

### Categorie

Scopri di più su Ordinary Differential Equations in Help Center e File Exchange

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by