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

#### karush

##### Well-known member
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
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

ok here is the example I am trying to follow

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#### topsquark

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

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

#### topsquark

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

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

Exactly right.

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
the derivatives are different with trig functions
I think anyway

#### topsquark

##### Well-known member
MHB Math Helper
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
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.