MHB -17.2.0 Solve y''+7y'+10y=80 by undetermined coefficients.

[karush]

$\tiny{17.2.0}$
$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\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]

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]

$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\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]

$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\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:

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

Okay, now what form will the particular solution take?

 

[karush]

\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 $
$y_p^´=0$
$y_p^{´´}=0$

 

[MarkFL]

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:

$$y''+7y'+10y=80$$

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:

$$y_p(x)=A$$

The particular solution will be a constant.

And thus:

$$y_p'(x)=0$$

$$y_p''(x)=0$$

Substituting into the ODE, what do we obtain?

 

[karush]

$10y=80$
$y=10$

 

[MarkFL]

$10y=80$
$y=10$

Upon substitution, we have:

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

And so we have:

$$y_p(x)=8$$

And so our general solution is:

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

And there's our 8...:D

 

[karush]

I'm beside myself

 

[HallsofIvy]

If you think \frac{80}{10}= 10 you certainly are.