# Question about the def. of solving 2nd order ODEs through Var. of Parameters.

• Je m'appelle
In summary, the method of variation of parameters assumes that c1 and c2 are variables and that y1 and y2 are solutions to the 2nd order ODE with the same initial conditions. By substituting u1 and u2 into the equation for c1y1 and c2y2, the particular solution for yp can be found.
Je m'appelle
Ok, so I've been studying the method of variation of parameters in order to solve 2nd order ODEs, and I have a question regarding a supposition that is made in the definition of the method.

Say,

$$y'' + p(t)y' + q(t)y = g(t)$$

Then the general solution to the above equation is

$$c_1y_1(t) + c_2y_2(t) + y_p$$, and now replacing $$c_1\ and\ c_2\ by\ u_1(t)\ and\ u_2(t)$$ on the complementary solution $$c_1y_1(t) + c_2y_2(t)$$ in order to find the particular solution $$y_p$$

$$y_p = u_1(t)y_1(t) + u_2(t)y_2(t)$$

$$y'_p = u'_1(t)y_1(t) + u_1(t)y'_1(t) + u'_2(t)y_2(t) + u_2(t)y'_2(t)$$

And then the methods states that

$$u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0$$

And this is where I'm curious and confused at the same time, I would like to know why it can be considered true, is there some kind of a proof to back this up? I mean, the person who developed this method certainly had something in mind to state that, so what does it consist of?

Then we get to derive $$y'_p$$ once again and then we substitute the values for $$y''_p,\ y'_p\ y_p$$ on the original equation to get the following system

$$u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0\ OBS:\ I\ wanna\ know\ why\ this\ is\ equal\ to\ zero.$$

$$u'_1(t)y'_2(t) + u'_2(t)_2y'_2(t) = g(t)$$

So we can solve for $$u_1(t)\ and\ u_2(t)$$ and find the particular solution for the ODE.

Leaving us with the general solution for the ODE

$$y = c_1y_1(t) + c_2y_2(t)\ -\ y_1(t)\int \frac{y_2(t)g(t)}{W(y_1,y_2)(t)}\ dt\ +\ y_2(t)\int \frac{y_1(t)g(t)}{W(y_1,y_2)(t)}\ dt$$

Thanks.

Last edited:
You have defined two functions $u_i$, and you only have one constraint on them, which is your differential equation. You are free to choose your second constraint however you wish, and the constraint you're asking about happens to be convenient.

Je m'appelle said:
Ok, so I've been studying the method of variation of parameters in order to solve 2nd order ODEs, and I have a question regarding a supposition that is made in the definition of the method.

Say,

$$y'' + p(t)y' + q(t)y = g(t)$$

Then the general solution to the above equation is

$$c_1y_1(t) + c_2y_2(t) + y_p$$, and now replacing $$c_1\ and\ c_2\ by\ u_1(t)\ and\ u_2(t)$$ on the complementary solution $$c_1y_1(t) + c_2y_2(t)$$ in order to find the particular solution $$y_p$$

$$y_p = u_1(t)y_1(t) + u_2(t)y_2(t)$$

$$y'_p = u'_1(t)y_1(t) + u_1(t)y'_1(t) + u'_2(t)y_2(t) + u_2(t)y'_2(t)$$

And then the methods states that

$$u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0$$

And this is where I'm curious and confused at the same time, I would like to know why it can be considered true, is there some kind of a proof to back this up? I mean, the person who developed this method certainly had something in mind to state that, so what does it consist of?

It's hard to guess what the original developer of the method was thinking. It might have been as easy as "since cty1+c2y2 doesn't have enough flexibility to work, maybe if I let the c's be variables, I can make it work". But keep in mind that the goal is to see if somehow a u1(t) and u2(t) can be found that will work. Given that these are two unknown functions that you hope to solve for you need solving for them not to be as difficult or more so than the original DE. The author cleverly noted that by setting that equation equal to zero, he can avoid getting any second derivatives of the u's when they are substituted into the second order DE. It isn't that he noted that the above equation must be zero. Rather, it was an assumption that was made to try to keep solving for the u's as simple as possible. So it turns out that the system of equations that come out for the u derivatives can be solved and, in principle, the u functions themselves and hence yp.

Surely he must have shouted "Yes!" and celebrated with a glass of beer!

## 1. What is the definition of a second order ODE?

A second order ordinary differential equation (ODE) is a mathematical equation that describes how a function changes over time, based on the values of the function and its first and second derivatives. It is written in the form: y'' + p(x)y' + q(x)y = g(x), where y is the unknown function, p and q are known functions of x, and g is a known function of x.

## 2. What does it mean to solve a second order ODE?

Solving a second order ODE means finding a function y(x) that satisfies the equation y'' + p(x)y' + q(x)y = g(x) for a given range of values of x. This function is called the solution of the ODE.

## 3. What is the method of variation of parameters for solving second order ODEs?

The method of variation of parameters is a technique used to find the solution of a second order ODE. It involves assuming a solution of the form y = u(x)y1(x) + v(x)y2(x), where y1 and y2 are two linearly independent solutions of the corresponding homogeneous ODE. The functions u(x) and v(x) are then determined by substituting this solution into the original ODE and solving for their derivatives.

## 4. When is the method of variation of parameters used to solve second order ODEs?

The method of variation of parameters is used when the coefficients p(x) and q(x) in the ODE are not constant, and therefore cannot be solved using other methods such as the method of undetermined coefficients or the method of reduction of order.

## 5. What are the advantages and disadvantages of using the method of variation of parameters?

The main advantage of the method of variation of parameters is that it can be used to find the particular solution of a nonhomogeneous second order ODE, even when the coefficients are not constant. However, it is a more complicated and time-consuming method compared to other techniques, and may not be suitable for all types of ODEs.

• Calculus and Beyond Homework Help
Replies
2
Views
434
• Calculus and Beyond Homework Help
Replies
7
Views
689
• Calculus and Beyond Homework Help
Replies
3
Views
601
• Calculus and Beyond Homework Help
Replies
3
Views
730
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
9
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
965