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I have a differential equation which I think is pretty much unsolvable analytically and numerically, as I do not know $y(c_0)$. $y(c_0)$ is the value of $y(x)$ at $c_0$, however I do not know this value.

$y'(x)=x y(x)^n+\frac{1}{x}(y(c_0)-y(x))^{-q}-\alpha y(x)$

Can I deduce from this equation that $y(x)$ will be decreasing in $\alpha$? If yes, why? Can I even deduce whether $y(x)$ is decreasing or increasing in $x$?

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  • $\begingroup$ What does $y(c_0)$ represent, is it related to an initial condition? $\endgroup$ – WalterJ Aug 2 at 9:24
  • $\begingroup$ Basically yes, but unfortunately I do not know the value, but $y'(x)$ depends on $y(c_0)$ $\endgroup$ – Paul Aug 2 at 9:37
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If there are other specifications that allow you to show that $y'$ is always positive or negative for all $x$, then you can infer that $y$ Is monotone increasing or decreasing with $x$. But to determine how $y'$ changes with $\alpha$, you'd need to know the magnitude of the two other terms relative to $\alpha y$.

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Just for the sake of experimentation, this MATLAB code integrates some version of your equation (note, I took $y(c_0)=y_0$). For this example $y\to 0$, but this is not generally true!

%% ODE example
clear all; clc; close all; 
n=2;
q=-1;
a=-2;
t0 = 1;
t1 = 100; 
tspan = [t0 t1];
y0 = -100;
[t,y] = ode23s(@(t,y) t*(y^n)+(1/t)*(y0-y)^(-q)-a*y, tspan, y0);
plot(t,y,'-k')
grid on 
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  • 1
    $\begingroup$ Thank you Amy and walter $\endgroup$ – Paul Aug 2 at 14:40

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