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

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Patrick Guarente
Patrick Guarente 2017 年 9 月 25 日
回答済み: Lewis Fer 2021 年 6 月 10 日
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 件のコメント
Patrick Guarente
Patrick Guarente 2017 年 9 月 25 日
Yes the analytical solution in terms of A and B.


回答 (4 件)

Teja Muppirala
Teja Muppirala 2017 年 9 月 26 日
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
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.

Lewis Fer
Lewis Fer 2021 年 6 月 10 日
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;
the boundary conditions are: f'(0)=0 et h'(o)=5;
g(0)=3 et h'(2)=h(2)

James Tursa
James Tursa 2017 年 9 月 25 日
編集済み: James Tursa 2017 年 9 月 25 日
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.
>> 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 件のコメント
Patrick Guarente
Patrick Guarente 2017 年 9 月 25 日
I'm not sure where those two equations will be implemented.


Torsten 2017 年 9 月 26 日
According to MATHEMATICA, the analytical solution is
z(x) = log(cosh(sqrt(A*B)*x))/A
Best wishes

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