# Undetermined coefficients

• February 6th 2009, 10:03 AM
jazz836062
Undetermined coefficients
I am just wanting somebody to check my work.

5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). $y''-2y'+y=7e^t cos(t)$

The auxiliary equation I get is:
$r^2-2r+1=0$
$(r-1)^2=0$
$r=1$ Twice

So,
$\alpha = \frac2 2 = 1$
$\beta = \sqrt(4-4)/2) = 0$
That means that s=1

So the equation I get is:
$yparticular(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)" alt="yparticular(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)" />
which simplifies to
$y(t)=A*t*e^t$
• February 6th 2009, 10:06 AM
Jester
Quote:

Originally Posted by jazz836062
I am just wanting somebody to check my work.

5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). $y''-2y'+y=7e^t cos(t)$

The auxiliary equation I get is:
$r^2-2r+1=0$
$(r-1)^2=0$
$r=1$ Twice

So,
$\alpha = \frac2 2 = 1$
$\beta = \sqrt(4-4)/2) = 0$

And $\alpha \;\text{and}\; \beta$ are?
• February 6th 2009, 10:22 AM
jazz836062
Quote:

Originally Posted by danny arrigo
And $\alpha \;\text{and}\; \beta$ are?

In my book the method of undetermined Coefficients calls for $\alpha$ and $\beta$.

$\alpha = \frac{-b} {2a}$
$\beta = \frac{\sqrt{4ac-b^2}} {2a}$

The plug it into the equation:

$y_{particular}(t)=t^s(A_{m}t^m+...+A_{0})e^{\alpha t}cos(\beta t)+t^s(B_{m}t^m+...+B_{0})e^{\alpha t}sin(\beta t)$

where m=degree of t in original equation.
s=0 if $\alpha + i \beta$ is not a root
s=1 if $\alpha + i \beta$ is a root
• February 6th 2009, 10:37 AM
Jester
Quote:

Originally Posted by jazz836062
I am just wanting somebody to check my work.

5. Find the Particular solution of the following equation using Undetermined Coefficients(yparticular(t). $y''-2y'+y=7e^t cos(t)$

The auxiliary equation I get is:
$r^2-2r+1=0$
$(r-1)^2=0$
$r=1$ Twice

So,
$\alpha = \frac2 2 = 1$
$\beta = \sqrt(4-4)/2) = 0$
That means that s=1

So the equation I get is:
$y_p(t)=t*A*t^0*e^t*cos(0*t)+t*B*t^0*e^t*sin(0*t)$
which simplifies to
$y(t)=A*t*e^t$

Quote:

Originally Posted by jazz836062
In my book the method of undetermined Coefficients calls for \alpha and \beta.

$\alpha = \frac{-b} {2a}$
$\beta = \frac{\sqrt{4ac-b^2}} {2a}$

The plug it into the equation:

$y_{particular}(t)=t^s(A_{m}t^m+...+A_{0})e^{\alpha t}cos(\beta t)+t^s(B_{m}t^m+...+B_{0})e^{\alpha t}sin(\beta t)$

where m=degree of t in original equation.
s=0 if $\alpha + i \beta$ is not a root
s=1 if $\alpha + i \beta$ is a root

Instead of memorizing formula's, you could do the following.

First, the complementary $y''-2y'+y=0$. As you said, the roots are 1, 1 so

$y_c = c_1 e^t + c_2 t e^t$

For the particular, ask is the nonhomogenous part contained in the complementary solution - no. So seek a particular solution of the form

$y_p = A e^t \cos t + B e^t \sin t$

Substitute gives

$-A e^t \cos t - B e^t \sin t = 7 e^t \cos t$

Comparing gives A = -7 and B = 0 so

$y_p = -7 e^t \cos t$

and the solution

$y = y_c + y_p = c_1 e^t + c_2 t e^t -7 e^t \cos t$
• February 6th 2009, 01:06 PM
jazz836062
Thank you. I got it wrong on the test. I was going back and checking my answers after the fact and i got it way wrong. Thank you for your help.