variation of parameters: y'' + y = sin(x) + x cos(x)

[rooney]

y''+y=sin(x)+xcos(x)

I need help finding the variables for the special function. in my book they are called u1 and u2. in my text it tells us to find u1' and u2' using wronskians involving the right hand side and y1 and y2 from the homogeneous equation, but it has no examples of a RHS with more than one function. do i carry on as usual or is there an extra step involved? sorry if this is unclear. i'll try to explain more if need be.

 

[galactus]

Re: variation of parameters

Let's step through it.

\(\displaystyle y''+y=sin(x)+xcos(x)\)

The auxiliary equation is \(\displaystyle m^{2}+1=0, \;\ m=i, \;\ m=-1\)

Since the solutions to the auxiliary equation are complex, we have:

\(\displaystyle y_{c}=C_{1}cos(x)+C_{2}sin(x)\)

\(\displaystyle y_{1}=cos(x), \;\ y_{2}sin(x)\)

\(\displaystyle f(x)=sin(x)+xcos(x)\)

\(\displaystyle W(cos(x),sin(x))=\begin{vmatrix}cos(x)&sin(x)\\-sin(x)&cos(x)\end{vmatrix}=1\)

\(\displaystyle W_{1}=\begin{vmatrix}0&sin(x)\\sin(x)+xcos(x)&cos(x)\end{vmatrix}=-xsin(x)cos(x)-sin^{2}(x)\)

\(\displaystyle W_{2}=\begin{vmatrix}cos(x)&0\\-sin(x)&sin(x)+xcos(x)\end{vmatrix}=xcos^{2}(x)+cos(x)sin(x)\)

\(\displaystyle u_{1}^{'}=\frac{W_{1}}{W}=-xsin(x)cos(x)-sin^{2}(x)\)

\(\displaystyle u_{2}^{'}=\frac{W_{2}}{W}=xcos^{2}(x)+cos(x)sin(x)\)

Now, integrate to get \(\displaystyle u_{1}, \;\ u_{2}\):

\(\displaystyle u_{1}=\frac{sin(x)cos(x)-2xsin^{2}(x)-x}{4}\)

\(\displaystyle u_{2}=\frac{2xsin(x)cos(x)+x^{2}-cos^{2}(x)}{4}\)

\(\displaystyle y_{p}=cos(x)\left(\frac{sin(x)cos(x)-2xsin^{2}(x)-x}{4}\right)+sin(x)\left(\frac{2xsin(x)cos(x)+x^{2}-cos^{2}(x)}{4}\right)\)

Believe it or not, that big mess above reduces to \(\displaystyle y_{p}=\frac{x^{2}}{4}sin(x)-\frac{x}{4}cos(x)\)

Therefore, \(\displaystyle \boxed{y=y_{c}+y_{p}=C_{1}cos(x)+C_{2}sin(x)+\frac{x^{2}}{4}sin(x)-\frac{x}{4}cos(x)}\)

 

[rooney]

Re: variation of parameters

Thanks a ton.