- #1
matematikuvol
- 192
- 0
Eq
[tex]u'(x)+p(x)u=f(x)[/tex]
with initial condition [tex]u(0)=0[/tex]
It's homogenous solution is
[tex]u_h=Ce^{-\int^x_0 p(s)ds}[/tex]
Complete solution
[tex]u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds=\int^x_0 f(s)g(x,s)ds[/tex]
where [tex]g(x,s)=e^{-\int^{x}_s p(\xi)d \xi }[/tex]
I didn't see that last step from here [tex]u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds[/tex] to [tex]\int^x_0 f(s)g(x,s)ds[/tex] here.
Can you explain me this?
[tex]u'(x)+p(x)u=f(x)[/tex]
with initial condition [tex]u(0)=0[/tex]
It's homogenous solution is
[tex]u_h=Ce^{-\int^x_0 p(s)ds}[/tex]
Complete solution
[tex]u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds=\int^x_0 f(s)g(x,s)ds[/tex]
where [tex]g(x,s)=e^{-\int^{x}_s p(\xi)d \xi }[/tex]
I didn't see that last step from here [tex]u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds[/tex] to [tex]\int^x_0 f(s)g(x,s)ds[/tex] here.
Can you explain me this?