Question

I'm wondering if somebody can check my algebra on an ODE separation of variables problem. Posted below shortly.

Answer 1

\[xy''-(x+1)y'+y=x^2e^x; \ \ \ y_{1}(x)=e^{x}, \ x>0.\]

Answer 2

First, verifying that the given solution is also a solution to the homogeneous variant of this DE: \[y_{1}=e^x; \ \ \ y_1 '=e^x; \ \ \ y_1 '' = e^x. \]\[xe^x-(x+1)e^x+e^x=0. \]The solution is valid; this solution can therefore be used to find another.

Answer 3

Let y_1{x}=ve^x; \[y=ve^x;\ \ \ y- = v'e^x+ve^x; \ \ \ y'' = v''e^x+2v'e^x+ve^x\]

Answer 4

Second expression should be equal to y', not y-.

Answer 5

\[xy'-(x+1)y'+y=x^2e^x; \ \ \ \]\[xe^x(v''+2v'+v)-(x+1)e^x(v'+v)+v=x^2e^x\]

Answer 6

\[v''x+(x-1)v'=x^2\]

Answer 7

@wio , if this is correct, I know I'm somehow supposed to put this in the form of a first order linear ODE in order to use an integrating factor, but I'm not sure how.

Answer 8

@Directrix

Answer 9

I think you can make a substitution at that last step. We know how to easily solve first order linear ODE--as you had suggested, we can multiply by an integrating factor. From your last step:
\[\text{Let w = v'}\]
\[w'x + (x-1)w = x^2\]
Now put it in standard form:
\[w' + \frac{x-1}{x}w = x\]
Which is now a first order linear ODE which should look more familiar.

Answer 10

Thank you!

Answer 11

In dealing with this technique with second order ODE's, should there be an assumption that-if the technique works with the problem-all zeroth order forms of your dependent variable disappear? e.g. you should never get a regular v in the problem, only v'', v', and so on?

Answer 12

Long story short, I'm working on a separate problem, and after the substitution w where you replace a 2nd order derivative with a 1st order one, I am getting: \[w'+a(x)w+\int\limits_{}^{}w=f(x)\]
This clearly can't be right, so I'm guessing an algebraic error was made (I found it), and in the way this works, is it fair to say that you can not have any zeroth order terms originally for your substitution to work?

Answer 13

e.g., after your first substitution in any given reduction of order problem where the technique is guaranteed to work, if you end up with the form \[v''a(x)+v'b(x)+vc(x)=f(x), \]something *must* have been done wrong, because using the substitution \[w'=v''\] will result in one of the terms having to be integrated (the zeroth order v term).

Answer 14

Looking at the problem again, aren't you supposed to use this method to find the homogeneous solution, then a different method (perhaps variation of parameters) to solve for the nonhomogeneous solution?

Regardless, I think the first degree term (i.e. the "y" term as opposed to the y' or y'' terms) disappears because of this:

For the homogeneous equation
\[y'' + P(x) y' + Q(x) y = 0\]
we make the assumption that a second solution y_2 can be given based off of the given solution, y_1:
\[y = u(x) y_1\]
Making things easier, I rewrite it as
\[y = uy_1\]
Then the first and second derivatives would be given as
\[y' = uy_1' + y_1 u'\]
\[y'' = uy_1'' + 2u'y_1' + u''y_1\]
Plugging into the homogeneous equation:
\[\left( uy_1'' + 2u'y_1' + u''y_1 \right) + P \left( uy_1' + y_1u' \right) + Q \left( uy_1 \right) = 0\]
This can be rearranged to
\[u \left( y_1'' + Py_1' + Qy_1 \right) + Py_1 u' + u'' y_1 + 2u'y_1' = 0\]
Where the term in the parentheses is just the homogeneous equation (equals 0). That's why the u term always disappears.