Solving Differential Equation Using Reduction of Order

[giodude]

Hello,
I'm working on the following differential equation:In problem 23 through 30 use the method of reduction of order to find a second solution of the given differential equation.
27. xy'' - y' + 4*x^3*y = 0, x > 0; y_1(x) = sin(x^2)
I begin my solution by writing out the known equations of y, y', y'' in terms of v and y_1. I also differentiate y_1 to find y_1', y_1'':
Let v denote v(x):
y = v*y_1
y' = v'*y_1 + v*y_1'
y'' = v''*y_1 + 2v'y_1' + v*y_1''

y_1 = sin(x^2) (given)
y_1' = 2*x*cos(x^2)
y_1'' = 2*cos(x^2) - 4*x^2*sin(x^2)

Next, I plug y, y', and y'' into the original differential:
x*(v''*y_1 + 2*v'*y_1' + v*y_1'') - v'*y_1 - v*y_1' + 4*x^3*v*y_1 = 0
x*v''*y_1 + (2*x*y_1' - y_1)*v' + (xy'' - y' + 4*x^3*y)v = 0 (the coefficient of v is our original differential so we may set that to 0)
x*v''*y_1 + (2*x*y_1' - y_1)*v' = 0
v'' + (2*x*y_1' - y_1) / (x*y_1)*v' = 0

Now we need to solve this first order differential equation (I use integration factors to solve):
d[mu]/dx = (2*x^2*cos(x^2)/xsin(x^2) - sin(x^2)/(x*sin(x^2))) * mu
d[mu]/dx = (2*x*cot(x^2) - 1/x)*mu
(1/mu)*d[mu]/dx = 2*x*cot(x^2) - 1/x

Differentiating both sides:
ln(mu) = -ln(sin(x^2)) - ln(x)

Exponentiating both sides:
mu = 1/(x*sin(x^2))

Plugging the integrating factor back in we get:
d[(1/(x*sin(x^2))) * v']/dx = 0
1/(x*sin(x^2)) * v' = C_1
v' = C_1*x*sin(x^2)

Integrating v':
v = -C_1*(1/2)*cos(x^2) + C_2

We can collapse -C_1*(1/2) into a C_1 to represent that entire constant term. Solving for y by computing v*y_1:
y = C_1*cos(x^2)*sin(x^2) + C_2*sin(x^2)

The answer in the textbook I'm using says that y_2 = cos(x^2), however this indicates a solution of y_2 = cos(x^2)*sin(x^2). Is this an interpretation problem such that this answer does in fact suggest y_2 = cos(x^2) or did I go wrong somewhere within my process of solving?

Thank you!

 

[pasmith]

I don't agree with your result for [itex]v[/itex].

Your equation for [itex]v'[/itex] can be reduced to [tex]
\begin{split}
0 &= xy_1v'' + (2xy_1' - y_1)v' \\
&= \frac{x^2}{y_1} \left( \frac{y_1^2}x v'' + \left( \frac{2y_1y_1'}{x} - \frac{y_1^2}{x^2}\right)v'\right) \\
&= \frac{x^2}{y_1} \frac{d}{dx}\left( \frac{y_1^2 v'}{x}\right) \end{split}[/tex] from which it follows that [tex]v' = \frac{Cx}{y_1^2} = \frac{Cx}{\sin^2(x^2)}.[/tex] Comparing this to your working suggests that something has gone wrong in your application of the integrating factor method, so I would suggest you try that again.

My equation for [itex]v'[/itex] can be integrated by substitution, leading to the book's answer.

 

[giodude]

Differentiating both sides:
ln(mu) = -ln(sin(x^2)) - ln(x)

Oh, found my two mistakes:
(1) I dropped the two when integrating 2*x*cot(x^2).
(2) I added a negative (mixed up with tan(x^2)) when integrating cot(x^2).

Fixing these two leads to the proper answer. Thank you!