Solving second order unhomogonous equations

[sweep123]

Homework Statement

solve y"+y=xsin(x) , initial conditions y(0)=0, y'(0)=1

Homework Equations

The Attempt at a Solution

I know that when right hand side equals something like sin(x), then the particular integral is xc1sin(x)+xc2cos(x). I am unsure what the particular integral should be for above. can anyone help me. I have tried (x^2)c1sin(x)+(x^2)c2cos(x), but have not gotten a solution from it. Thanks

 

[Dick]

You need a bigger family of trial functions. Try c1*x^2*sin(x)+c2*x^2*cos(x)+c3*x*sin(x)+c4*x*cos(x).

 

[Dmak]

instead of looking at that equation you could try to look at an equivalent equation
$$L[y] = y'' + y = xe^{ix}$$, where $$i = \sqrt{-1}$$. then since $$i$$ is a particular root of the characteristic equation, y has a solution of the form
$$y = x^{2}A_{0}e^{ix}$$ from there it is easily shown that the complex part of the solution of $$y$$ is the solution of the original equation you proposed.

You could also try the variation of parameters method but I often avoid it because of messy integration

 

[Dick]

instead of looking at that equation you could try to look at an equivalent equation
$$L[y] = y'' + y = xe^{ix}$$, where $$i = \sqrt{-1}$$. then since $$i$$ is a particular root of the characteristic equation, y has a solution of the form
$$y = x^{2}A_{0}e^{ix}$$ from there it is easily shown that the complex part of the solution of $$y$$ is the solution of the original equation you proposed.

You could also try the variation of parameters method but I often avoid it because of messy integration

I really don't know what you are talking about, but neither the real nor the imaginary part of x^2*exp(ix) is a particular solution to y''+y=x*sin(x).

 

[Dmak]

I really don't know what you are talking about, but neither the real nor the imaginary part of x^2*exp(ix) is a particular solution to y''+y=x*sin(x).

whoops what i meant was a particular solution of the form:
$$y = x(A_{0} + A_{1}x)e^{xi}$$ you have to derive the coefficients by evaluating $$L[y]$$ and matching up the coefficients $$A_{0}, A_{1}$$ with the right hand side of the equation ( $$xe^{ix}$$ ) then you use the imaginary portion of this solution, this one was more tedious than I thought and I would recommend using the variation of parameters method.

 

[sweep123]

thanks. that's really helpful. is there anyway of knowing what the particular integral is for any unhomogenous second order differential equation, or is it just trial and error?

 

[sweep123]

ive just tried substituting the particular integral
y(x)= c1(x^2)sin(x)+c2(x^2)cos(x)+xc3sin(x)+xc4cos(x) into y"+y=xsinx. I end up with
2c1sin(x)+4xc1cos(x)+2c2cos(x)-4xc2sin(x)+2c3cos(x)-2c4sin(x)=xsin(x)

This doesn't look solvable. I've checked my differentiating a couple of times and it looks okay. Any suggestions where I may be going wrong?

 

[Dmak]

This would be a fine method: http://www.sosmath.com/diffeq/second/variation/variation.html

essentially find two homogeneous solutions $$y_{1}, y_{2}$$ then using those you can find a solution to the differentiall equation of the form,

$$\phi(x) = y_{1}u_{1} + y_{2}u_{2}$$
where
$$u'_{1} = \frac{-sin(x)y_{2}}{W[y_{1}, y_{2}]}$$
$$u'_{2} = \frac{sin(x)y_{1}}{W[y_{1}, y_{2}]}$$
and
$$W[y_{1}, y_{2}] = y_{1}'y_{2} - y_{1}y_{2}'$$

 

[Dick]

ive just tried substituting the particular integral
y(x)= c1(x^2)sin(x)+c2(x^2)cos(x)+xc3sin(x)+xc4cos(x) into y"+y=xsinx. I end up with
2c1sin(x)+4xc1cos(x)+2c2cos(x)-4xc2sin(x)+2c3cos(x)-2c4sin(x)=xsin(x)

This doesn't look solvable. I've checked my differentiating a couple of times and it looks okay. Any suggestions where I may be going wrong?

It is solvable. How about c1=c4=0 and c2=-1/4 and c3=1/4? You just have to try harder.

 

[sweep123]

oh yeah. i should have spent a bit more time on that. Thanks