-17.2.01 Solve the given equation by the method of undetermined coefficients

[karush]

$\tiny{17.2.01}\\$
$\textrm{ Solve the given equation by the method of undetermined coefficients.}$
\begin{align*}\displaystyle
y''+7y'+10y&=80
\end{align*}
$\textit{this is in the form}$
\begin{align*}\displaystyle
x^2+7x+10&=0\\
(x+2)(x+5)&=0
\end{align*}
$\textit{this is a second-order linear ordinary differential equation so}$
\begin{align*}\displaystyle
y& = c_1 e^{-5 x} + c_2 e^{-2 x}+8
\end{align*}

ok I know stuff is missing here?

 

[MarkFL]

You look at what's on the RHS and see it is a constant, and you also note that no term of the homogeneous solution is a constant, therefore you assume the particular solution must be a constant:

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

And so:

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

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

Substituting $y_p$ into the ODE, we obtain:

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

Add so we have:

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

Now, using the principle of superposition, we find the solution to the ODE to be:

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

 

[karush]

think I got it
let me try another one.

my class here is over so it will the Summer of Review..

and a peek thru the door of Calc III

of which I have heard horror stories

 

[MarkFL]

think I got it
let me try another one.

my class here is over so it will the Summer of Review..

and a peek thru the door of Calc III

of which I have heard horror stories

When I was in school, my classmates were mixed about which was more difficult, Calc II or Calc III...I personally thought Calc II was a bit more difficult of the two. :)