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
Quote:
Originally Posted by
FernandoRevilla Yes, you have proved that the given function is a solution of the differential equation.
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.
The DE...
$\displaystyle x^{2}\ y^{''} - x\ y^{'} + 2\ y=0$ (1)
... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...
$\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)
...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...
$\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)
Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Re: 2nd order differential equation problem
Quote:
Originally Posted by
chisigma The DE...
$\displaystyle x^{2}\ y^{''} - x\ y^{'} + 2\ y=0$ (1)
... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...
$\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)
...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...
$\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)
Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
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...
$\displaystyle x^{2}\ y^{''} - x\ y^{'} + 2\ y=0$ (1)
... is 'Euler's type' and a particular solution of it has the form $\displaystyle x^{\alpha}$. Substituting $\displaystyle x^{\alpha}$ in (1) You arrive at the algebraic equation...
$\displaystyle \alpha^{2}-2\ \alpha +2 =0$ (2)
...the solution of which are $\displaystyle 1 \pm i$, so that the general solution of (1) is...
$\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (3)
Of course if You want a real solution it must be $\displaystyle c_{1}=c_{2}=c$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
My answer has been a little hurried so that what i wrote is not precise(Doh)... the general solution of the De is...
$\displaystyle y(x)=c_{1}\ x\ (\cos \ln x +i\ \sin \ln x) + c_{2}\ x\ (\cos \ln x -i\ \sin \ln x)$ (1)
... bue $\displaystyle c_{1}$ and $\displaystyle c_{2}$ may be complex, so that the general real solution is...
$\displaystyle y(x)= x\ (a_{1}\ \cos \ln x + a_{2}\ \sin \ln x)$
... where $\displaystyle a_{1}$ and $\displaystyle a_{2}$ are 'arbitrary real constants'...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
