MHB -m30b Convert the differential equation

[karush]

ov!347 nmh{1000}

Convert the differential equation

$$y''+5y'+6y=e^x$$
into a system of first order (nonhomogeneous) differential equations and solve the system.

the characteristic equation is
$$\lambda^2+5\lambda+6=e^x$$
factor
$$(\lambda+2)(\lambda+3)=e^x$$

ok not real sure what to do with this $=e^x$ thing

 

[topsquark]

Convert the differential equation

$$y''+5y'+6y=e^x$$
into a system of first order (nonhomogeneous) differential equations and solve the system.

the characteristic equation is
$$\lambda^2+5\lambda+6=e^x$$
factor
$$(\lambda+2)(\lambda+3)=e^x$$

ok not real sure what to do with this $=e^x$ thing

The characteristic equation is [math]\lambda ^2 + 5 \lambda + 6 = 0[/math]. You don't need the [math]e^x[/math] until later.

-Dan

 

[karush]

$(\lambda+2)(\lambda+3)=0$
the root are then
$\lambda-2, \quad \lambda-3$
so then we have
$e^{-2x},\quad e^{-3x} $

then hopefully

$y=c_1e^{-2x}+c_2e^{-3x}$

so how do we finish this ? with =e^x

 

[topsquark]

$(\lambda+2)(\lambda+3)=0$
the root are then
$\lambda-2, \quad \lambda-3$
so then we have
$e^{-2x},\quad e^{-3x} $

then hopefully

$y=c_1e^{-2x}+c_2e^{-3x}$

so how do we finish this ? with =e^x

I was commenting on what the characteristic equation is, not what you are supposed to do for this problem.

Let's define A(x) = y(x) and B(x) = y'(x). Then
A'(x) = y'(x)
B'(x) = y''(x)

Putting this into your original equation gives
[math]y'' = -5y' - 6y + e^x[/math]

or
[math]B' = -5B - 6A + e^x[/math]

And don't forget: [math]A' = B[/math], from the original definition of B.

So you have the system of differential equations:
[math]\left ( \begin{matrix} A' \\ B' \end{matrix} \right ) = \left ( \begin{matrix} B \\ -6A - 5B \end{matrix} \right )+ \left ( \begin{matrix} 0 \\ e^x \end{matrix} \right )[/math]

or, using the usual notation:
[math]\left ( \begin{matrix} A \\ B \end{matrix} \right ) ^{\prime} = \left ( \begin{matrix} 0 & 1 \\ -6 & -5 \end{matrix} \right ) \left ( \begin{matrix} A \\ B \end{matrix} \right ) + \left ( \begin{matrix} 0 \\ e^x \end{matrix} \right )[/math]

Now you have a pair of simultaneous first order linear differential equations, which you've been studying.

-Dan

 

[HallsofIvy]

Convert the differential equation

$$y''+5y'+6y=e^x$$
into a system of first order (nonhomogeneous) differential equations

This first part has not been addressed. Let z= y'. Then y''= z'.
Then $z'+ 5z+ 6y= e^z$ or $z'= e^x- 5z- 6y$. Together with z= y' we have the two equations y'= z and $z'= e^x- 5z- 6y$.

That could also be written as the matrix equation
$\begin{pmatrix}y \\ z \end{pmatrix}'= \begin{pmatrix}0 & 1 \\ -6 & -5 \end{pmatrix}\begin{pmatrix}y \\ z \end{pmatrix}+ \begin{pmatrix} 0 \\ e^x\end{pmatrix}$.

The characteristic equation for the differential equation is the characteristic equation for that matrix, $\left|\begin{array}{cc}-\lambda & 1 \\ -6 & -5- \lambda\end{array}\right|= 0$.