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

In summary, the mathematician is trying to find the solution to a problem, but is lost and needs help from a friend. The friend helps by providing a solution to the problem.
  • #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?
 
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  • #2
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:
[tex]\int_0^x f(s) \exp\left\{ \int_0^s p(z) dz - \int_0^x p(z) dz \right\} \, ds [/tex]
(note I renamed the dummy integration variable to z).

Now you can rewrite the integral in the exponent to
[tex]\int_0^s p(z) dz - \int_0^x p(z) dz = \int_x^s p(z) dz[/tex]
and just call the whole thing g(x, s).
 
  • #3
Tnx. Just to ask you how I get this boundaries in integration?

[tex]u_h'+p(x)u_h=0[/tex]

[tex]\frac{du_h}{dx}=-p(x)u_h(x)[/tex]

[tex]u_h=Ce^{-\int p(x)dx}[/tex]

So

[tex]u(x)=C(x)e^{-\int p(x)dx}[/tex]

[tex]u'=C'(x)e^{-\int p(x)dx}-pu[/tex]

When I put this to Eq

[tex]u'+p(x)u=f(x)[/tex]

I get

[tex]C'(x)e^{-\int p(x)dx}-pu+pu=f[/tex]

and

[tex]C(x)=\int f(x)e^{\int p(x)dx}dx + C'[/tex]

So solution is

[tex]u(x)=(\int f(x)e^{\int p(x)dx}dx + C')e^{-\int p(x)dx}[/tex]

How to get integration border [tex][a,b][/tex], [tex]\int^{b}_a[/tex] in integrals. Tnx for your help.
 
  • #4
Can you help me?
 
  • #5


Sure, I'd be happy to explain. The last step is essentially just simplifying the expression using the definition of the function g(x,s) which was given in the problem. Let's break it down step by step:

1. Starting with u(x)=e^{-\int^x_0 p(s)ds}\int^x_0f(s)e^{\int^s_0 p(z)dz}ds, we can use the property of exponents to rewrite it as u(x)=\int^x_0f(s)e^{\int^s_0 p(z)dz-\int^x_0 p(s)ds}ds.

2. Next, we can use the definition of g(x,s) to rewrite the exponent as \int^s_0 p(z)dz-\int^x_0 p(s)ds=\int^s_0 p(z)dz+\int^s_0 p(s)ds=\int^s_0 p(z)+p(s)dz.

3. Substituting this back into the original expression, we get u(x)=\int^x_0f(s)e^{\int^s_0 p(z)+p(s)dz}ds.

4. Using the property of integrals, we can rewrite the exponent as \int^x_0 p(z)+p(s)dz=\int^x_0 p(z)dz+\int^x_0 p(s)dz.

5. Finally, we can use the definition of g(x,s) again to rewrite the integral as \int^x_0 p(s)dz=\int^x_0 p(s)g(x,s)ds.

Combining all of these steps, we get u(x)=\int^x_0 f(s)g(x,s)ds, which is the final expression given in the problem. I hope this explanation helps clarify the last step for you.
 

1. What is a homogeneous equation?

A homogeneous equation is a mathematical equation in which all terms have the same degree. This means that the coefficients of each term are raised to the same power.

2. What are initial conditions in the context of solving homogeneous equations?

Initial conditions refer to the values of the unknown function and its derivatives at a specific point. These values are used to find a unique solution to the homogeneous equation.

3. How do you solve a homogeneous equation with initial conditions?

To solve a homogeneous equation with initial conditions, you can use the method of separation of variables. This involves rewriting the equation in terms of two separate functions and then using the initial conditions to find the constant of integration.

4. What is the role of the integral in the equation u(x)=\int^x_0 f(s)g(x,s)ds?

The integral in this equation represents the solution to the original homogeneous equation. It allows us to express the function u(x) in terms of the given functions f(s) and g(x,s).

5. Can you provide an example of solving a homogeneous equation with initial conditions?

For example, we can solve the equation y'' + 4y' + 3y = 0 with initial conditions y(0) = 1 and y'(0) = -2. After using the method of separation of variables, we get the solution y(x) = 2e^{-x} - e^{-3x}. Plugging in the initial conditions, we get the final solution y(x) = e^{-x} - e^{-3x}.

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