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[SOLVED] 17.2.0 Solve the given equation by the method of undetermined coefficients.

karush

Well-known member
Jan 31, 2012
2,928
$\tiny{17.2.0}$
$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\begin{align*}\displaystyle
y''+7y'+10y&=80\\
\end{align*}
$\textrm{auxilary equation}$
\begin{align*}\displaystyle
r^2+7r+10&=0\\
(r+2)(r+5)&=0\\
r&=-2,-5
\end{align*}

then is it the general equation?

this is the last one of the undetermined coefficients problems
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You have correctly identified the auxiliary roots...so:

a) What is the homogeneous solution $y_h$?

b) What form will the particular solution $y_p$ take?
 

karush

Well-known member
Jan 31, 2012
2,928
$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\begin{align*}\displaystyle y''+7y'+10y&=80\\\end{align*}
$\textrm{auxilary equation}$
\begin{align*}\displaystyle
r^2+7r+10&=0\\
(r+2)(r+5)&=0\\
r&=-2,-5
\end{align*}
$y_h=c^{-5x}+c^{-2x}+8$
$y_p=Ax^2+Bx + 8$
I think?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\begin{align*}\displaystyle y''+7y'+10y&=80\\\end{align*}
$\textrm{auxilary equation}$
\begin{align*}\displaystyle
r^2+7r+10&=0\\
(r+2)(r+5)&=0\\
r&=-2,-5
\end{align*}
$y_h=c^{-5x}+c^{-2x}+8$

I think?
What we would have is:

\(\displaystyle y_h(x)=c_1e^{-5x}+c_2e^{-2x}\)

Okay, now what form will the particular solution take?
 

karush

Well-known member
Jan 31, 2012
2,928
\begin{align*}\displaystyle y''+7y'+10y&=80\\\end{align*}
$\textrm{auxilary equation}$
\begin{align*}\displaystyle
r^2+7r+10&=0\\
(r+2)(r+5)&=0\\
r&=-2,-5
\end{align*}
$y_h=c_1 e^{-5x}+c_2 e^{-2x}$
$y_p=A $
$y_p^´=0$
$y_p^{´´}=0$
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Let's not get ahead of ourselves. :D To determine the form for the particular solution, we look at the RHS of the original ODE:

\(\displaystyle y''+7y'+10y=80\)

We see it (80) is a constant, and we also note that no term in the homogeneous solution is a constant, and so the particular solution takes the form:

\(\displaystyle y_p(x)=A\)

The particular solution will be a constant.

And thus:

\(\displaystyle y_p'(x)=0\)

\(\displaystyle y_p''(x)=0\)

Substituting into the ODE, what do we obtain?
 

karush

Well-known member
Jan 31, 2012
2,928
$10y=80$
$y=10$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$10y=80$
$y=10$
Upon substitution, we have:

\(\displaystyle 0+7(0)+10A=80\implies A=8\)

And so we have:

\(\displaystyle y_p(x)=8\)

And so our general solution is:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1e^{-5x}+c_2e^{-2x}+8\)

And there's our 8...:D
 

karush

Well-known member
Jan 31, 2012
2,928
I'm beside myself😎
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
If you think [tex]\frac{80}{10}= 10[/tex] you certainly are.