Can Variation of Parameters Solve This First-Order Differential Equation?

[Hammie]

I'm not currently in a class, but I'm doing this for fun.. but technically I would still call it coursework, so I'm posting it here..

I'm studying Redheffer/Sokolnikoff's Mathmatics of Modern Engineering.. and I find a problem on page 75, use the method of variation of parameters to find the solution of this equation:

Y` + 3Y = X^3

this one is easy enough to find the forced response by plugging in
Y= AX^3 + BX^2 + CX + D = X^3, and equating the coefficients.

The coefficients turn out to be A=1/3, B=-1/3, C=-2/9, D=-2/27.

This is the answer, but that's not my problem.

The variation of parameter method discussed in this section requires two linearly independent solutions to the homogenous equation,, of which I can only find one:

Y= Ke^(-3t).

Of course there is the trivial solution, but that would make the Wronksian in the denominator of the integrals zero.

For the life of me, I can't see how in the world one can use the variation of parameters method to solve this..

Is it perhaps a trick question? I'd email Sokolnikoff.. but he's long since left the world..



any ideas?

 

[Galileo]

In the method of variation of parameters you let the constant(s) of integration be functions of the independent variable. In this case there is only one such constant (since it's a first order D.E.), so let y(x)=K(x)exp(-3x).

Now plug it in your differential equation and solve for K.

 

[Hammie]

This is what this section is calling the method of variation of parameters:

$$Y(t) = -y_1 \int \frac{y_2 f}{W(y_1,y_2)}dt + y_2\int\frac{y_1 f}{W(y_1,y_2)}dt$$
What I'm trying to figure out.. is this asking for something that cannot be done?

You still need two linearly independent solutions Y1 and Y2 to the homogenous equation, along with only the forcing function f for it to work. In this problem, f = x^3.

I am beginning to think that putting a first order DE in this problem set was a mistake..

 

[Galileo]

Well, can't say that expression looks familiar...

But you won't be able to find two linearly independent solutions to the homogeneous equation. There's only one, since it's first order and you found it: Kexp(-3x)

Variation of parameters is where you t let the arbitrary constants in the solution to the homogeneous equation be functions of the independent variable in order to find the general solution.

Did a search and found this. So it's the same thing, but your expression only works for 2nd-order equations. My advice: just let y(x)=K(x)exp(-3x) and plug it in the equation. It'll work out perfectly, but you'll be solving an annoying integral with lots of integration by parts.

 

[Hammie]

But you won't be able to find two linearly independent solutions to the homogeneous equation. There's only one, since it's first order and you found it: Kexp(-3x)

right. That's kind of the point.. in this particular book, I just couldn't figure why it was asking me to compute a driven response with two linearly independent solutions to the homogenous equation, when I could only come up with one solution.

your reference shows the same thing, this method is for second order equations.

That may be a common thing in these old books.. it was published in 1958..

"do so and so"

answer is, "can't do it that way.."

 

[HallsofIvy]

If that was really a first order equation : Y'+3Y= x³
rather than a second order equation, Y"+ 3y= x³, then there are not two independent solutions!

But "variation of parameters" works for any degree linear equation. The formula you wrote is just the result of substuting y= K(x)Y₁ + L(x)Y_{2[/sup] (Y1 and Y2 are solutions to the homogeneous equation). In the first order case it reduces to what Galileo said: Let y= K(x)e^-3x and substitute into the equation. You will get a simple equation for K'.}