Simple differential equation substitution

[Jamp]

Hi!

I am looking through some solved exercises. One of them is the following:

Solve the equation: x^2 y'' + (x^2 - 3x)y' + (3-x)y = x^4
knowing that y=x is a solution of the homogeneous equation.

The professor then solves it by doing the following substitution: y=xz.
Then he calculates y' and y'', substitutes, etc.

What I do not understand is, how do you know, from the fact that y=x is a solution to the homogeneous equation, that you have to do that substitution y=xz?

 

[J77]

What is z?

 

[HallsofIvy]

Hi!

I am looking through some solved exercises. One of them is the following:

Solve the equation: x^2 y'' + (x^2 - 3x)y' + (3-x)y = x^4
knowing that y=x is a solution of the homogeneous equation.

The professor then solves it by doing the following substitution: y=xz.
Then he calculates y' and y'', substitutes, etc.

What I do not understand is, how do you know, from the fact that y=x is a solution to the homogeneous equation, that you have to do that substitution y=xz?

You don't "know" that- you don't have to do that substitution- there are many ways to solve a differential equation. However, it is true that what ever function the correct solution y is, there exist a function z such that y= xz: z= y/x, of course.

The point is this: if y= f(x)z (for ANY function f(x) that satisfies the equation), then y'= f'(x)z+ f(x)z', y"= f"(x)z+ 2f'(x)z'+ f(x)z". I've used the product rule twice here. Notice that, in the terms that involve only z, not z' or z", I have differentiated only the f(x) term. It is exactly as if z were a constant. Since f(x) satisfies this linear homogenous equation so does any constant times f(x): the terms of involving only z wil all cancel out leaving a differential equation for z' and higher derivatives. Now let u(x)= z' and you have a differential equation for u of lower order.

To repeat: you don't have to do that substitution- but it can be done and has a good chance of simplifying the problem so it is worth trying!

 

[Jamp]

I see.
So if I were told that y=x^2 was a solution to the homogeneous eq, then I should try y=x^2z?

 

[HallsofIvy]

Yes, exactly. That's known as "reduction of order": x^2 is a solution to an nth order linear homogeneous diff eq, then y= x^2z will be a solution for some z satisfying an n-1 order equation. It's very similar to reducing the degree of a polynomial by dividing by x-a if you already know that a is a solution.

 

[rbzima]

Just solve it using a power series. It's far easier to do that then to try and find substitutions every single time.

 

[HallsofIvy]

Just solve it using a power series. It's far easier to do that then to try and find substitutions every single time.

What "substitutions" are you talking about? The question here was about using the fact that you know one solution to a differential equation to reduce the order.

 

[rbzima]

I'm saying that if you know y = xz, then for every variable in the equation you are trying to substitute to reduce to a single variable. While that is helpful, it doesn't always yield a solution to the equation. Therefore, in instances when you don't know y = xz, it would be wise to solve using a power series.

Perhaps its just the fact that I love power series that it would come more natural for me to do so in that manner.

 

[dextercioby]

Hi!

I am looking through some solved exercises. One of them is the following:

Solve the equation: x^2 y'' + (x^2 - 3x)y' + (3-x)y = x^4
knowing that y=x is a solution of the homogeneous equation.

The professor then solves it by doing the following substitution: y=xz.
Then he calculates y' and y'', substitutes, etc.

What I do not understand is, how do you know, from the fact that y=x is a solution to the homogeneous equation, that you have to do that substitution y=xz?

The method of variation of constants. $y_{p}(x)=C(x) y_{o}(x)=C(x) x=z(x)x=zx$.