2nd order differential equation problem

I just wanted to confirm I did this right.

"Verify by substitution that the given function is a solution of the given differential equation."

Given:

x^2 y'' - xy' + 2y = 0

y = x*cos( ln(x) )

My work:

y' = cos(ln(x)) - sin(ln(x))

y'' = -(1/x) * ( sin(ln(x)) + cos(ln(x)) )

Substituting back in to the differential equation:

x^2 (-(1/x)(sin(ln(x)) + cos(ln(x)))) - x(cos(ln(x)) - sin(ln(x))) - 2(xcos(ln(x)) = 0

-xsin(ln(x)) - xcos(ln(x)) - xcos(ln(x)) + xsin(ln(x)) + 2x*cos(ln(x)) = 0

0 = 0

Does this mean I verified the solution? See I'm confused because in first order diff. equations I would just separate out the x's and y's and integrate both sides and get my y equation. This is a little different apparently.

Re: 2nd order differential equation problem

Quote:

Originally Posted by

**nautica17** Does this mean I verified the solution?

Yes, you have proved that the given function is a solution of the differential equation.

Quote:

See I'm confused because in first order diff. equations I would just separate out the x's and y's and integrate both sides and get my y equation. This is a little different apparently.

The order of the equation is irrelevant. One thing is to solve a differential equation and another thing is to verify that a __given__ function is a solution.

Re: 2nd order differential equation problem

Re: 2nd order differential equation problem

Quote:

Originally Posted by

**chisigma** The DE...

(1)

... is 'Euler's type' and a particular solution of it has the form

. Substituting

in (1) You arrive at the algebraic equation...

(2)

...the solution of which are

, so that the general solution of (1) is...

(3)

Of course if You want a real solution it must be

...

Kind regards

This confirms that it is not the same to find the general solution of a DE that verify that a given function is a solution of a DE.

Re: 2nd order differential equation problem

Quote:

Originally Posted by

**chisigma** The DE...

(1)

... is 'Euler's type' and a particular solution of it has the form

. Substituting

in (1) You arrive at the algebraic equation...

(2)

...the solution of which are

, so that the general solution of (1) is...

(3)

Of course if You want a real solution it must be

...

Kind regards

My answer has been a little hurried so that what i wrote is not precise(Doh)... the general solution of the De is...

(1)

... bue and may be complex, so that the general real solution is...

... where and are 'arbitrary real constants'...

Kind regards