-m30 - 2nd order linear homogeneous ODE solve using Wronskian

[karush]

2000
Convert the differential equation
$$\displaystyle y^{\prime\prime} + 5y^\prime + 6y =0$$
ok I presume this means to find a general solution so
$$\lambda^2+5\lambda+6=(\lambda+3)(\lambda+2)=0$$
then the roots are
$$-3,-2$$
thus solutions
$$e^{-3x},e^{-2x}$$
ok I think the Wronskain matrix is next which I have never done.

well so far,,,, any suggest...
also this might be DE question ... not sure..

 

[Ackbach]

Sorry, I really have no idea what your question is. You've solved the DE correctly (well, you could write $y=c_1 e^{-2x}+c_2 e^{-3x},$ but you've done all the hard work), what else is there to do? Not sure what the problem statement means by "convert". Does the book define that term?

 

[karush]

View attachment 9118

ok here is the example I am trying to follow

 

[topsquark]

Convert the differential equation
$$\displaystyle y^{\prime\prime} + 5y^\prime + 6y =0$$
ok I presume this means to find a general solution so
$$\lambda^2+5\lambda+6=(\lambda+3)(\lambda+2)=0$$
then the roots are
$$-3,-2$$
thus solutions
$$e^{-3x},e^{-2x}$$
ok I think the Wronskain matrix is next which I have never done.

well so far,,,, any suggest...
also this might be DE question ... not sure..

Well, the Wronskian is
[math]W = \left | \begin{matrix} f(x) & g(x) \\ f'(x) & g'(x) \end{matrix} \right |[/math]
with [math]f(x) = e^{-3x}[/math] and [math]g(x) = e^{-2x}[/math]

If [math]W \neq 0[/math] then the two functions f(x) and g(x) are linearly independent. (Actually when [math]W \neq 0[/math] on some interval [math]x \in I[/math] of the reals then f and g are independent on that interval.)

What do you get?

-Dan

 

[karush]

well far ...

$\displaystyle
W=\left|\begin{matrix}e^{3x}&e^{2x}\\-3 e^{3 x}& -2 e^{2 x}\end{matrix}\right|\\
=(e^{3x})(2 e^{-2 x})-(3 e^{3 x})( e^{2 x})\\
=2 e^{5 x}-3 e^{5x}
=-e^{5x}$

 

[topsquark]

well far ...

$\displaystyle
W=\left | \begin{matrix}e^{-3x}&e^{-2x} \\-3 e^{-3 x} & -2 e^{-2 x} \end{matrix} \right |$

not sure

Good! Now calculate the determinant.

-Dan

 

[karush]

I think you can drop the negatives ... (oh wait the original!)

anyway I updated the previous post...

thot I could slip under the wire$W=\left | \begin{array}{cc} e^{- 3 x} & e^{- 2 x} \\ - 3 e^{- 3 x} & - 2 e^{- 2 x} \end{array} \right |=e^{- 5 x}$

 

[topsquark]

I think you can drop the negatives ... (oh wait the original!)

anyway I updated the previous post...

thot I could slip under the wire$W=\left | \begin{array}{cc} e^{- 3 x} & e^{- 2 x} \\ - 3 e^{- 3 x} & - 2 e^{- 2 x} \end{array} \right |=e^{- 5 x}$

Also good! Now, can W be 0 for any values of x?

-Dan

 

[karush]

$$e^{- 5 x}\ne 0$$
so $e^{-3x},e^{-2x}$
$\text{are linearly independent and form a fundamental set of solutions
So the general solution is}$

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

no book answer so hopefully...

 

[topsquark]

$$e^{- 5 x}\ne 0$$
so $e^{-3x},e^{-2x}$
$\text{are linearly independent and form a fundamental set of solutions
So the general solution is}$

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

no book answer so hopefully...

Exactly right. (Handshake)

Unless you need to use the Wronskian you can take it as a given that [math]e^{ax}[/math] and [math]e^{bx}[/math] are linearly independent for [math]a \neq b[/math]. What can you say about sin(ax) and cos(ax)?

-Dan

 

[karush]

the derivatives are different with trig functions
I think anyway

 

[topsquark]

the derivatives are different with trig functions
I think anyway

Yes, but the Wronskian is the same form:
[math]W = \left | \begin{matrix} sin(ax) & cos(ax) \\ a ~ cos(ax) & -a ~ sin(ax) \end{matrix} \right | [/math]

[math]= -a ~ sin^2(ax) - a ~ cos^2(ax)[/math]

[math] = - a \left ( sin^2 (ax) + cos^2 (ax) \right ) [/math]

[math]W = -a[/math]

Can this be zero for any values of a?

If you have some time work out if sin(ax) and cos(bx) are linearly independent.

-Dan

 

[HallsofIvy]

ok here is the example I am trying to follow

The word "convert" does not appear in that at all. It does not appear to have anything to do with your original post. To "convert" something means to "change" it to something else. What were you to convert this equation to?

Is this related to your subsequent post where you were to "convert" a second order differential equation to a system of first order equations?

If so, then letting z= y', z'= y'' and the second order equation y''+ 5y'+ 6y= 0 becomes z'+ 5z+ 6y= 0. The "system of first order equations" is
y'= z
z'= -5z- 6y.