Complementary function and particular integral

[songoku]

Homework Statement

The quantities x and y are related by the differential equation :

$$\frac{dy}{dx}+3y=6x+5$$

(i) Find the complementary function of th above differential equation
(ii) Find also a particular integral for this differential equation, and hence write down the general solution

Homework Equations

Integrating factor

The Attempt at a Solution

I know how to find the general solution, but I don't know what complementary function and particular integral are...

Thanks

 

[tiny-tim]

Hi songoku!

I know how to find the general solution, but I don't know what complementary function and particular integral are...

Complementary function (or complementary solution) is the general solution to dy/dx + 3y = 0.

Particular integral (I prefer "particular solution") is any solution you can find to the whole equation.

(basically, you make a wild but intelligent guess … in this case, I'd go for a polynomial )

Since the difference between any two particular solutions is a complementary function, that means that you can choose just one particular solution, add all the complementary functions to it, and you get all the solutions.​

 

[vela]

If you use an integrating factor to solve this problem, you end up with an integral of the integrating factor multiplied by (6x+5). The portion of the solution which comes from this term, sans the integration constant, might be what's being called the particular integral.

 

[songoku]

Hi tiny-tim and vela

Complementary function (or complementary solution) is the general solution to dy/dx + 3y = 0.

So, differential equation will have complementary solution only if the form :
dy/dx + (a)y = r(x) ?

Particular integral (I prefer "particular solution") is any solution you can find to the whole equation.

particular integral = particular solution?

This is what I read on my book :
"the general solution of a differential equation, involving an arbitrary constant, represents more than one, even infinitely many, solutions. the solutions obtained by assigning definite value to the constant is called a particular solution. typically, the value is determined by making the solution satisfy a given condition, traditionally referred to as initial condition."

So I think particular integral is not the same as particular solution. To find particular solution, we have to find the general solution first.

(basically, you make a wild but intelligent guess … in this case, I'd go for a polynomial )

how to make wild but intelligent , such that you can guess polynomial? and how to apply it to solve the question?

my book never mentioned about complementary function and particular integral

If you use an integrating factor to solve this problem, you end up with an integral of the integrating factor multiplied by (6x+5). The portion of the solution which comes from this term, sans the integration constant, might be what's being called the particular integral.

I found the general solution : y = (6x+5)/3 - 2/3 + Ce^(-3x). You mean the particular integral is (6x+5)/3 - 2/3 ?

Thanks ^^

 

[vela]

This is what I read on my book :
"the general solution of a differential equation, involving an arbitrary constant, represents more than one, even infinitely many, solutions. the solutions obtained by assigning definite value to the constant is called a particular solution. typically, the value is determined by making the solution satisfy a given condition, traditionally referred to as initial condition."

So I think particular integral is not the same as particular solution. To find particular solution, we have to find the general solution first.

The general solution to a linear differential equation consists of two pieces. One piece is the solution to the homogeneous equation; the other is the solution to the inhomogeneous equation. The latter piece is typically referred to as the particular solution. Your book isn't using the term in this way; perhaps that is why the problem refers to a particular integral instead, to avoid confusion.

I found the general solution : y = (6x+5)/3 - 2/3 + Ce^(-3x). You mean the particular integral is (6x+5)/3 - 2/3?

Yes, but keep in mind I'm just making an educated guess. If u(x) is the integrating factor, the solution is given by

$$y=\frac{1}{u(x)}\int u(x)(6x+5)dx + \frac{C}{u(x)}$$

The term due to the arbitrary constant is the solution to the homogeneous equation. Since
the first term contains an integral and yields what's typically called the particular solution (though not by your book), I figured that's what particular integral means. So tiny-tim and I are saying the same thing: the particular integral is what most people call the particular solution.

 

[tiny-tim]

Hi songoku!

So, differential equation will have complementary solution only if the form :
dy/dx + (a)y = r(x) ?

No, the complementary solution is where the RHS is zero (the "homogenoeus equation").

particular integral = particular solution?

This is what I read on my book :
"the general solution of a differential equation, involving an arbitrary constant, represents more than one, even infinitely many, solutions. the solutions obtained by assigning definite value to the constant is called a particular solution. typically, the value is determined by making the solution satisfy a given condition, traditionally referred to as initial condition."

So I think particular integral is not the same as particular solution.

No, I think they're the same …

"the solution obtained by assigning a definite value to the constant" is a particular solution (in this case, it has the constant = 0).

Remember, a "particular solution" is any, repeat ANY, solution to the complete equation.​

To find particular solution, we have to find the general solution first.

No NO NOOOO!

You can find the particular solution first.

It usually has nothing to do with the general complementary solution (in this case, it was linear, and the complementary solution was exponential).

how to make wild but intelligent , such that you can guess polynomial? and how to apply it to solve the question?

I believe there are some rules somewhere, but I don't remember them.

Usually, it's fairly obvious … you look at the RHS, and imitate it …

in this case, the RHS is a polynomial, so you try a polynomial.

I found the general solution : y = (6x+5)/3 - 2/3 + Ce^(-3x). You mean the particular integral is (6x+5)/3 - 2/3 ?

Yes, except you'll lose marks in the exam if you don't simplify that, to 2x +1.

 

[songoku]

Hi vela and tiny-tim

The general solution to a linear differential equation consists of two pieces. One piece is the solution to the homogeneous equation; the other is the solution to the inhomogeneous equation. The latter piece is typically referred to as the particular solution. Your book isn't using the term in this way; perhaps that is why the problem refers to a particular integral instead, to avoid confusion.

If so, what is the term for "particular solution", which is the solution we get using the initial value to find the constant? In my book, it is called particular solution, but if particular solution = particular integral, maybe you know other term for calling my definition of "particular solution"

( I hope I don't cause confusion...)

No, the complementary solution is where the RHS is zero (the "homogenoeus equation").

dy/dx + 3y = 6x + 5

dy/dx - 6x = 5 - 3y

Can I say complementary function is solution to :
dy/dx - 6x = 0 ?

No, I think they're the same …

"the solution obtained by assigning a definite value to the constant" is a particular solution (in this case, it has the constant = 0).

Remember, a "particular solution" is any, repeat ANY, solution to the complete equation.​

But my book said the value of the constant is obtained using the initial condition given from the question and the constant is not always zero.

Usually, it's fairly obvious … you look at the RHS, and imitate it …

in this case, the RHS is a polynomial, so you try a polynomial.

So if the RHS is exponential, I try exponential, and if it is trigonometry, I also try trigonometry?

Thanks a lot

 

[tiny-tim]

Hi songoku!

So if the RHS is exponential, I try exponential, and if it is trigonometry, I also try trigonometry?

Yup!

dy/dx + 3y = 6x + 5

dy/dx - 6x = 5 - 3y

Can I say complementary function is solution to :
dy/dx - 6x = 0 ?

oops! i should have been more specific …

the complementary solution is the general solution when the LHS has only functions of y (so not including functions of x, or constants)

to put it another way, y is there as (d/dx)⁰y, the 0th derivative of y

If so, what is the term for "particular solution", which is the solution we get using the initial value to find the constant? In my book, it is called particular solution, but if particular solution = particular integral, maybe you know other term for calling my definition of "particular solution"
…
But my book said the value of the constant is obtained using the initial condition given from the question and the constant is not always zero.

I suspect your book is asking eg "Find the particular solution (or integral) of dy/dx + 3y = 6x + 5 for which y = 7 when x = 12" …

this is using "particular" in its non-technical sense, just as one might ask "find the equation of the particular tangent to this circle that passes through (12,7)".

 

[songoku]

Hi tiny-tim

I suspect your book is asking eg "Find the particular solution (or integral) of dy/dx + 3y = 6x + 5 for which y = 7 when x = 12" …

this is using "particular" in its non-technical sense, just as one might ask "find the equation of the particular tangent to this circle that passes through (12,7)".

Actually for this question, it doesn't ask about particular solution. But the others are just like what you wrote (there is initial value to find the particular solution).

Let take y = 2 and x = 0 for this question, then the particular solution will be :

y = 2x + 1 + 1/(e^3x) and this is not the same as particular integral, which is y = 2x + 1

I think I understand how to find the complementary solution and particular integral. I just a little bit confused about the terms "particular solution" and "particular integral"

Thanks

 

[vela]

I searched the web a bit, and it seems there are two senses in how the term particular solution applies to differential equations. The more common one is in referring to the solution to the non-homogeneous equation. This solution is sometimes called the particular integral, as in your homework problem. But the term particular solution is also apparently used to draw a distinction between a general solution with still arbitrary constants and the solution in which those constants have specific values to satisfy the initial conditions. This latter usage is what your book uses.

To summarize: Your problem uses "particular" in the more common sense, and the particular integral is the same thing as the particular solution. You can ignore what your book says.

 

[tiny-tim]

But the term particular solution is also apparently used to draw a distinction between a general solution with still arbitrary constants and the solution in which those constants have specific values to satisfy the initial conditions. This latter usage is what your book uses.

ah! as in the (non-scientific) expression "From the general to the particular".

 

[songoku]

ah! as in the (non-scientific) expression "From the general to the particular".

ah! That's what I mean Thanks a lot to you both !

*NB
tiny-tim, can you help me on other thread about increasing function (you've answered the question earlier). Mark has different opinion..https://www.physicsforums.com/showthread.php?t=372401

tiny-tim and vela : please help me too at this : https://www.physicsforums.com/showthread.php?t=372400

And I am very sorry if this is inappropriate or I break the rules...Thanks in advanced