# Thread: Method of Undetermined Coefficients

1. ## Method of Undetermined Coefficients

Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

ok so
yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)

2. Originally Posted by ur5pointos2slo
Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

ok so
yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)

Is the confusing part how the answer is setup or do you not know how to go about finding $\displaystyle y_P$?

3. Originally Posted by dwsmith
Is the confusing part how the answer is setup or do you not know how to go about finding $\displaystyle y_P$?
I do not know how to find yp. I read the chapter in the book but of course it's no help.

4. Do you know what annihilates $\displaystyle x*e^{\frac{-x}{2}}+x*cos{\frac{x}{2}}$?

5. Originally Posted by dwsmith
Do you know what annihilates $\displaystyle x*e^{\frac{-x}{2}}+x*cos{\frac{x}{2}}$?
no idea

6. $\displaystyle \bigg(D+\frac{1}{2}\bigg)^2\bigg(D^2+\frac{1}{4}\b igg)^2$

7. Originally Posted by dwsmith
$\displaystyle \bigg(D+\frac{1}{2}\bigg)^2\bigg(D^2+\frac{1}{4}\b igg)^2$
I still do not understand what is going on. Where did that come from?

8. Originally Posted by ur5pointos2slo
Find yc(complimentary solution) and set up yp(particular solution) by the method of undetermined coefficients.

y''+y'+(1/2)y = x*e^(-x/2)+x*cos(x/2)

ok so
yc = c1 e^(-x/2) cos(x/2) + c2 e^(-x/2) sin(x/2)

Finding yp is confusing me. Could someone please explain how to come to the answer. What assumptions must be made?

yp =(Ax + B)e^(-x/2) + (Cx+D)cos(x/2) + (Ex + F) sin(x/2)
There should be a standard table in your text that outlines what you should guess given a particular L.H.S

In this case,

$\displaystyle L.H.S = xe^{-x/2} + xcos(x/2)$

But let us note that we can use the prinipal of superposition here so

$\displaystyle L.H.S_1 = xe^{-x/2}$

$\displaystyle L.H.S_2 = xcos( \frac{x}{2} )$

Finally,

$\displaystyle L.H.S= L.H.S_1 + L.H.S_2$

So lets find an appropriate $\displaystyle Y_P$ for our L.H.S

For $\displaystyle L.H.S_1$ we note that we have the form

$\displaystyle f(x)e^{g(x)}$

Which by standard tables suggest we try the form

$\displaystyle Y_{P1} = (Ax+B)e^{g(x)}$

For $\displaystyle L.H.S_2$ we note that we have the form

$\displaystyle f(x)cos( g(x) )$

Which, again by stanard tables suggests we try the form

$\displaystyle Y_{P2} = (Cx + D)cos( p(x) ) + (Ex+F) sin ( p(x) )$

We include both Sin and Cos here because both are potential solutions (remember our trig circles!). Everytime you have sin or cos as your left hand side, you guess the appropriate function for both sin and cos.

In the end,

$\displaystyle Y_P= Y_{P1} + Y_{P2} = (Ax+B)e^{g(x)} (Cx + D)cos( p(x) ) + (Ex+F) cos ( p(x) )$

$\displaystyle (Ax+B)e^{ \frac{-x}{2} } (Cx + D)cos( \frac{x}{2} ) + (Ex+F) cos ( \frac{x}{2} )$

EDIT....I went through this entire thing use L.H.S when in fact we are using the R.H.S but it is late so dont castrate me lol

9. Originally Posted by AllanCuz
There should be a standard table in your text that outlines what you should guess given a particular L.H.S

In this case,

$\displaystyle L.H.S = xe^{-x/2} + xcos(x/2)$

But let us note that we can use the prinipal of superposition here so

$\displaystyle L.H.S_1 = xe^{-x/2}$

$\displaystyle L.H.S_2 = xcos( \frac{x}{2} )$

Finally,

$\displaystyle L.H.S= L.H.S_1 + L.H.S_2$

So lets find an appropriate $\displaystyle Y_P$ for our L.H.S

For $\displaystyle L.H.S_1$ we note that we have the form

$\displaystyle f(x)e^{g(x)}$

Which by standard tables suggest we try the form

$\displaystyle Y_{P1} = (Ax+B)e^{g(x)}$

For $\displaystyle L.H.S_2$ we note that we have the form

$\displaystyle f(x)cos( g(x) )$

Which, again by stanard tables suggests we try the form

$\displaystyle Y_{P2} = (Cx + D)cos( p(x) ) + (Ex+F) sin ( p(x) )$

We include both Sin and Cos here because both are potential solutions (remember our trig circles!). Everytime you have sin or cos as your left hand side, you guess the appropriate function for both sin and cos.

In the end,

$\displaystyle Y_P= Y_{P1} + Y_{P2} = (Ax+B)e^{g(x)} (Cx + D)cos( p(x) ) + (Ex+F) cos ( p(x) )$

$\displaystyle (Ax+B)e^{ \frac{-x}{2} } (Cx + D)cos( \frac{x}{2} ) + (Ex+F) cos ( \frac{x}{2} )$

EDIT....I went through this entire thing use L.H.S when in fact we are using the R.H.S but it is late so dont castrate me lol
lol thanks a lot.