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[SOLVED] m30 - 2nd order linear homogeneous ODE solve using Wronskian

karush

Well-known member
Jan 31, 2012
2,886
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..
 
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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,198
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

Well-known member
Jan 31, 2012
2,886
30.PNG

ok here is the example I am trying to follow
 
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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
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
\(\displaystyle W = \left | \begin{matrix} f(x) & g(x) \\ f'(x) & g'(x) \end{matrix} \right |\)
with \(\displaystyle f(x) = e^{-3x}\) and \(\displaystyle g(x) = e^{-2x}\)

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

What do you get?

-Dan
 

karush

Well-known member
Jan 31, 2012
2,886
well far ...

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

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
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

Well-known member
Jan 31, 2012
2,886
I think you can drop the negatives .... (oh wait the original!!!!!)

anyway I updated the previous post...

thot I could slip under the wire


$W=\left | \begin{array}{cc} e^{- 3 x} & e^{- 2 x} \\ - 3 e^{- 3 x} & - 2 e^{- 2 x} \end{array} \right |=e^{- 5 x}$
 
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topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
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

Well-known member
Jan 31, 2012
2,886
$$e^{- 5 x}\ne 0$$
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

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
$$e^{- 5 x}\ne 0$$
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 \(\displaystyle e^{ax}\) and \(\displaystyle e^{bx}\) are linearly independent for \(\displaystyle a \neq b\). What can you say about sin(ax) and cos(ax)?

-Dan
 

karush

Well-known member
Jan 31, 2012
2,886
the derivatives are different with trig functions
I think anyway
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,181
the derivatives are different with trig functions
I think anyway
Yes, but the Wronskian is the same form:
\(\displaystyle W = \left | \begin{matrix} sin(ax) & cos(ax) \\ a ~ cos(ax) & -a ~ sin(ax) \end{matrix} \right | \)

\(\displaystyle = -a ~ sin^2(ax) - a ~ cos^2(ax)\)

\(\displaystyle = - a \left ( sin^2 (ax) + cos^2 (ax) \right ) \)

\(\displaystyle W = -a\)

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

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
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.