On 08/24/2012 06:02 PM, Bertold Kraessig wrote:
> Hi,
>
> I am trying to solve a set of coupled 1st order differential equations
> which include terms of the type f(y)*y(x) with Maxima 5.26.0:
>
> test: 'diff(y(x),x)+f(x)*y(x)=a;
> atvalue(y(x),x=0,1);
> desolve(test,y(x));
>
> The result I am getting is:
>
> y(x)=ilt((( laplace( f(x)y(x), x, g1990 )-1) g1990 a)/g1990^2 , g1990, x )
>
> or, in the case of the specific function f(x):=exp(-x^2), a:0:
>
> y(x)= ilt ( ( laplace( exp(-x^2)y(x), x, g2042 ) -1 )/g2042 , g2042, x )
>
> How do I go from here to find the solution?
>
> In the latter case, Mathematica, returns the solution
> y(x)=exp( -sqrt(%pi)/2*erf(x) ),
> and in the former an expression involving exponentials of
> integrals of f(x).
>
Hi Bertold,
I apologized for taking almost 2 months to read your message, but in
case it is still useful, consider this:
(%i1) display2d: false$
(%i2) test: 'diff(y,x)+f(x)*y = a;
(%o2) 'diff(y,x,1)+f(x)*y = a
(%i3) ode2 (test,y,x);
(%o3) y = %e^-('integrate(f(x),x))
*(a*'integrate(%e^'integrate(f(x),x),x)+%c)
(%i4) subst ([a=0, f(x)=exp(-x^2)], %);
(%o4) y = %c*%e^-'integrate(%e^-x^2,x)
(%i5) ev (%,integrate);
(%o5) y = %c*%e^-(sqrt(%pi)*erf(x)/2)
(%i6) ic1 (%, x=0, y=1);
(%o6) y = %e^-(sqrt(%pi)*erf(x)/2)
Regards,
Jaime