2nd order ODE - Show solution by substitution

In summary, the conversation discusses a problem from a textbook and the attempt to solve it using integration by parts. However, it is suggested to use Leibnitz's rule to differentiate instead. The conversation also includes a helpful link to learn more about the rule.
  • #1
wxstall
7
0

Homework Statement



Show that

y(t) = (1/w) ∫[0,t] f(s)*sin(w(t-s)) ds

is a particular solution to

y'' +w2 y = f(t)where w is a constant.

The Attempt at a Solution



After wasting several pages of paper I have made virtually no progress. Obviously, substitution suggests you plug in y(t), differentiate it twice for the first term, and somehow arrive at f(t) = f(t). However, without more information about f(s), it seems impossible. Integration by parts on y(t) will result in another integral which in turn must be integrated by parts...and so on infinitely many times. Therefore integration by parts of the solution y(t) is not an option.

I have considered that this could be an application of the fundamental theorem of calculus, but with an additional sine term that also depends on t in the integrand, it seems not not apply (at least not in a way I am familiar with).This is a problem from Richard Haberman's text: Applied PDE with Fourier Series and Boundary Value Problems.

Any ideas/advice would be much appreciated.
Thank you
 
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  • #2
wxstall said:

Homework Statement



Show that

y(t) = (1/w) ∫[0,t] f(s)*sin(w(t-s)) ds

is a particular solution to

y'' +w2 y = f(t)


where w is a constant.

The Attempt at a Solution



After wasting several pages of paper I have made virtually no progress. Obviously, substitution suggests you plug in y(t), differentiate it twice for the first term, and somehow arrive at f(t) = f(t). However, without more information about f(s), it seems impossible. Integration by parts on y(t) will result in another integral which in turn must be integrated by parts...and so on infinitely many times. Therefore integration by parts of the solution y(t) is not an option.

I have considered that this could be an application of the fundamental theorem of calculus, but with an additional sine term that also depends on t in the integrand, it seems not not apply (at least not in a way I am familiar with).


This is a problem from Richard Haberman's text: Applied PDE with Fourier Series and Boundary Value Problems.

Any ideas/advice would be much appreciated.
Thank you

You don't need to integrate anything. You need to differentiate using Leibnitz's rule. There are several forms. You will find this form useful:$$
\frac d {dt} \int_0^t g(s,t)\, ds = \int_0^t \frac {\partial g(s,t)}{\partial t}\, ds + g(t,t)$$
 
  • #4
So as I suspected, it's some rule that I was unfamiliar with. Thanks for the help!
 

FAQ: 2nd order ODE - Show solution by substitution

1. What is a 2nd order ODE?

A 2nd order ODE (ordinary differential equation) is a mathematical equation that involves a function and its derivatives of the second order. It is commonly used in modeling physical systems and can be solved using various methods, such as substitution.

2. What is the process of solving a 2nd order ODE by substitution?

The process of solving a 2nd order ODE by substitution involves substituting the derivatives in the equation with new variables, such as u or v, to create a system of first order equations. These equations can then be solved using techniques like separation of variables or integrating factors.

3. When is it appropriate to use substitution to solve a 2nd order ODE?

Substitution is typically used to solve 2nd order ODEs when the equation is not in standard form, meaning the highest derivative is not isolated on one side of the equation. It can also be useful when the equation is nonlinear or difficult to solve using other methods.

4. What are the advantages of using substitution to solve a 2nd order ODE?

Substitution can simplify the process of solving a 2nd order ODE by breaking it down into a system of first order equations. It can also be used for equations that are not easily solved using other methods. Additionally, substitution can help to identify and classify different types of 2nd order ODEs.

5. Are there any limitations to using substitution to solve a 2nd order ODE?

While substitution can be a useful method for solving 2nd order ODEs, it may not always be the most efficient or accurate approach. In some cases, it may be more beneficial to use other techniques, such as Laplace transforms or power series methods. It is important to consider the specific equation and its properties before deciding on a method of solution.

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