Solving Homogenous Eq. with Initial Conditions: u(x)=\int^x_0 f(s)g(x,s)ds

[matematikuvol]

Eq

$$u'(x)+p(x)u=f(x)$$

with initial condition $$u(0)=0$$

It's homogenous solution is

$$u_h=Ce^{-\int^x_0 p(s)ds}$$

Complete solution

$$u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds=\int^x_0 f(s)g(x,s)ds$$

where $$g(x,s)=e^{-\int^{x}_s p(\xi)d \xi }$$

I didn't see that last step from here $$u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds$$ to $$\int^x_0 f(s)g(x,s)ds$$ here.

Can you explain me this?

 

[CompuChip]

I'll give you some clues, and leave the details up to you to figure out (if you don't manage, feel free to ask again).

You can take the prefactor into the integral, as it is constant w.r.t. s:
$$\int_0^x f(s) \exp\left\{ \int_0^s p(z) dz - \int_0^x p(z) dz \right\} \, ds$$
(note I renamed the dummy integration variable to z).

Now you can rewrite the integral in the exponent to
$$\int_0^s p(z) dz - \int_0^x p(z) dz = \int_x^s p(z) dz$$
and just call the whole thing g(x, s).

 

[matematikuvol]

Tnx. Just to ask you how I get this boundaries in integration?

$$u_h'+p(x)u_h=0$$

$$\frac{du_h}{dx}=-p(x)u_h(x)$$

$$u_h=Ce^{-\int p(x)dx}$$

So

$$u(x)=C(x)e^{-\int p(x)dx}$$

$$u'=C'(x)e^{-\int p(x)dx}-pu$$

When I put this to Eq

$$u'+p(x)u=f(x)$$

I get

$$C'(x)e^{-\int p(x)dx}-pu+pu=f$$

and

$$C(x)=\int f(x)e^{\int p(x)dx}dx + C'$$

So solution is

$$u(x)=(\int f(x)e^{\int p(x)dx}dx + C')e^{-\int p(x)dx}$$

How to get integration border $$[a,b]$$, $$\int^{b}_a$$ in integrals. Tnx for your help.

 

[matematikuvol]

Can you help me?