Method of Undetermined Coefficients

**ur5pointos2slo** · May 4th 2010, 06:30 PM

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?

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

**dwsmith** · May 4th 2010, 07:25 PM

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?

***ANSWER***
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 $y_P$ ?

**ur5pointos2slo** · May 4th 2010, 07:29 PM

Originally Posted by dwsmith

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

I do not know how to find yp. I read the chapter in the book but of course it's no help.

**dwsmith** · May 4th 2010, 07:31 PM

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

**ur5pointos2slo** · May 4th 2010, 07:34 PM

Originally Posted by dwsmith

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

no idea

**dwsmith** · May 4th 2010, 07:45 PM

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

**ur5pointos2slo** · May 4th 2010, 07:57 PM

Originally Posted by dwsmith

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

I still do not understand what is going on. Where did that come from?

**AllanCuz** · May 4th 2010, 08:59 PM

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?

***ANSWER***
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,

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

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

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

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

Finally,

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

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

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

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

Which by standard tables suggest we try the form

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

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

$f(x)cos( g(x) )$

Which, again by stanard tables suggests we try the form

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

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,

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

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

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

**ur5pointos2slo** · May 5th 2010, 04:13 AM

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,

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

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

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

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

Finally,

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

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

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

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

Which by standard tables suggest we try the form

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

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

$f(x)cos( g(x) )$

Which, again by stanard tables suggests we try the form

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

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,

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

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

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.

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