Please help me. I want to find the optimal values of constans in the ode?t

syms C1 C2 y(x)
dy = diff(y);
d2y = diff(dy);
eqn = y^2 * d2y + y*dy^2 - dy^2 + C1*y^4 + C2*y^3 == 0

eqn(x) = 

dsolve(eqn)

ans = 

[Torsten]

Optimal with respect to what ?

Answer 1

[Walter Roberson]

Provided that C1 and C2 are real-valued

syms y(x)
syms C1 C2 real
dy = diff(y);
d2y = diff(dy);
eqn = y^2 * d2y + y*dy^2 - dy^2 + C1*y^4 + C2*y^3 == 0

eqn(x) = 

sol = dsolve(eqn)

sol = 

Therefor the solutions to the equations are either

y = 0

or

y = -C2/C1

Provided that C1 is non-zero, these equations just are and there is no optimization to be done.

If C1 or C2 are non-real then the situation gets to be considerably messier.

[Umar]

@Torsten @Walter This exchange perfectly illustrates why clear problem statements matter in matlab answers! Torsten's question helped clarify that this was a DE solving problem, not optimization, and Walter's solution shows how elegant the actual answer is. The fact that such a complex-looking nonlinear ODE reduces to just two constant solutions is pretty remarkable.

[Sam Chak]

No @Umar, don't be misled by the 'dsolve' function for this seemingly straightforward ODE. The 'dsolve' solutions provided in @Tarek's post and @Walter Roberson's answer do not offer the complete picture. They are equilibrium points (algebraic solutions) where all derivative terms are zero.

I simplified the original ODE and created two cases to illustrate what a true analytical solution should look like. ODEs are equations in dynamical systems that describe the changes in y(x) with respect to changes in x. In most cases, if an analytical solution exists, it should be expressed in terms of x, and not static points. I seldom use the 'dsolve' function. Perhaps @Walter Roberson and @Torsten can shed some light on this matter.

For example, we expect the 1st-order linear system dy/dx + y(x) = 0 to give the exponential solution y(x) = y(0)·exp(-x).

syms y(x)
syms C1 C2 real
dy  = diff(y);
d2y = diff(dy);

%% Case 1
eqn1= y^2*d2y + C1*y^2 + C2*y == 0

eqn1(x) = 

sol1= dsolve(eqn1)

sol1 = 

%% Case 2 (when C2 is zero)
eqn2= y^2*d2y + C1*y^2 == 0

eqn2(x) = 

sol2= dsolve(eqn2)

sol2 = 

%% Solving Case 1 using ode45
function dy = case1(x, y)
    C1      = 1;
    C2      = 1;
    
    dy      = zeros(2, 1);
    dy(1)   = y(2);
    dy(2)   = - (C1*y(1)^2 + C2*y(1))/(y(1)^2);
end

init    = [1; 0];   % initial values: y(0) = 1 and y'(0) = 0
[x, y]  = ode45(@case1, [0 2], init);
plot(x, y), grid on
legend('y(x)', 'dy/dx', 'location', 'southeast', 'fontsize', 12)
title('The value of y(x) changes as x changes')
xlabel('x'), ylabel('y(x)')

[Umar]

After reviewing this exchange, I find myself agreeing with @Sam chai’s fundamental critique while appreciating the pedagogical value of this entire discussion.

@Walter Robertson’s analysis correctly identifies the equilibrium solutions, but Sam has exposed a critical limitation in how we interpret dsolve output for singular ODEs. The original equation:

y^2y’’ + y(y’)^2 - (y’)^2 + C1y^4 + C2y^3 = 0

is indeed a singular ODE due to the y^2 coefficient on the highest derivative term. When y=0, the equation degenerates, which explains why dsolve retreats to finding only the algebraic equilibrium points.

@Sam Chak’s numerical demonstration is particularly revealing. His phase portrait shows genuine dynamical evolution that the symbolic solver misses entirely. I could be wrong here but this highlights a broader issue in computational mathematics: symbolic solvers often provide incomplete solutions for nonlinear systems, especially those with degeneracies.

However, I would argue that both approaches have merit. The equilibrium analysis gives us the critical points of the dynamical system - essential for understanding stability and long-term behavior. @Sam Chak’s numerical approach reveals the transient dynamics between these equilibria.

The real lesson here is methodological: for singular nonlinear ODEs, we need a hybrid approach. Start with equilibrium analysis, then use numerical methods to understand the connecting trajectories. Neither method alone gives the complete picture.

This discussion perfectly illustrates that mathematical software requires critical interpretation, not blind trust.

[Torsten]

I think we didn't make much progress, did we ?

https://uk.mathworks.com/matlabcentral/answers/2178771-please-help-me-to-find-the-exact-solution-of-ode?s_tid=prof_contriblnk

Especially because @Tarek doesn't explain what he really aims at.

[Tarek]

Hi @Torsten.

Because the ODE : y^2y’’ + y(y’)^2 - (y’)^2 + C1y^4 + C2y^3 = 0 is difficult to find exact solution . Firstly,i want to find the exact solution of ODE any way using Matlab code.

If no code or method using Matlab to find the exact solution. I want to find the values of constants C1 and C2 such that the Numerical solution of of ODE is zero. Note B.C y(x=-any vale)=zero. y(x=+any value)=zero

[Torsten]

For that y == 0 is a solution of the ODE, you can choose arbitrary values for C1 and C2. In other words: Independent of C1 and C2, y == 0 is always a solution of the ODE.

[Walter Roberson]

We still need to know whether C1 and C2 are real-valued or not

[Tarek]

C1 and C2 are real values