## Help solving 2 ODEs

### Kurt (view profile)

on 24 Oct 2018
Latest activity Commented on by Star Strider

### Star Strider (view profile)

on 24 Oct 2018
Accepted Answer by Star Strider

### Star Strider (view profile)

W(0)=0; W(f)=2.8; d(P)/d(W)=((-0.3743)/(2*P))*(1-(0.15*X)); P(0)=1; d(X)/d(W)=0.5*((0.08*(0.75*(1-X)))/(1-(0.15*X))); X(0)=0; Trying to solve these two ODEs

R2018b

### Star Strider (view profile)

Answer by Star Strider

### Star Strider (view profile)

on 24 Oct 2018

Try this:
PX_ODE = @(W,PX) [((-0.3743)./(2*PX(1))).*(1-(0.15*PX(2))); 0.5*((0.08*(0.75*(1-PX(2))))./(1-(0.15*PX(2))))];
[W,PX] = ode15s(PX_ODE, [0, 2.8], [1; 0]);
figure
plot(W, PX)
grid
Here ‘P’ is ‘PX(1)’, ‘X’ is ‘PX(2)’. The system encounters a singularity at 2.67, and the integration stops.

Kurt

### Kurt (view profile)

on 24 Oct 2018
I mean to keep them in their original forms
Kurt

### Kurt (view profile)

on 24 Oct 2018
I was trying to do it this way, syms P(W) X(W) eqn1 = diff(P, W) == ((-0.3743)/(2*P))*(1-(0.15*X)); eqn2 = diff(X, W) == (0.5*((0.08*(0.75*(1-X)))/(1-(0.15*X))));
[odes, vars] = odeToVectorField(eqn1, eqn2); fun = matlabFunction(odes,'Vars',{'t','Y'}); X0(0) = [0, 0]; tspan = [0, 2.8]; [t, sol] = ode45(fun,tspan,x0);
P = sol(:,2); X = sol(:,1);
Star Strider

### Star Strider (view profile)

on 24 Oct 2018
That certainly works, although ‘P(0)’ cannot be zero. If ‘X(0)’ is greater than 1, the system will integrate out to 2.8.
Also, ‘fun’ is essentially the same as my code, and the result is the same. (The system is ‘stiff’, so ode15s or another stiff solver is more appropriate.)