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

• MHB
• karush
In summary, the conversation discusses converting a second order differential equation to a system of first order equations. The process involves finding a general solution by solving for the roots of the characteristic equation and then using the Wronskian to determine linear independence of the solutions. The conversation also touches on the linear independence of trigonometric functions and their derivatives.
karush
Gold Member
MHB
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..

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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?

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ok here is the example I am trying to follow

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karush said:
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

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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karush said:
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

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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karush said:
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

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

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

karush said:
$$e^{- 5 x}\ne 0$$
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. (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

the derivatives are different with trig functions
I think anyway

karush said:
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

karush said:
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.

## 1. What is a second order linear homogeneous ODE?

A second order linear homogeneous ODE (ordinary differential equation) is a mathematical equation that involves the second derivative of a function, as well as the function itself, and is equal to zero. It is called "linear" because the function and its derivatives appear in a linear fashion, and "homogeneous" because the equation is equal to zero.

## 2. What is the Wronskian method?

The Wronskian method is a technique for solving linear homogeneous ODEs that involves finding a particular solution using the Wronskian determinant, which is a mathematical tool that helps determine if a set of functions are linearly independent.

## 3. What does the -m30 parameter mean?

The -m30 parameter is a constant that is used in the Wronskian method to find a particular solution to the ODE. It is typically chosen based on the initial conditions of the problem and is used to simplify the Wronskian determinant.

## 4. How does the Wronskian method work?

The Wronskian method involves finding a particular solution to the ODE by using the Wronskian determinant to determine a set of linearly independent functions. These functions are then used to construct a general solution to the ODE, which can be further simplified by using the -m30 parameter.

## 5. What are the advantages of using the Wronskian method to solve ODEs?

The Wronskian method is advantageous because it is a systematic and efficient way to solve linear homogeneous ODEs. It also allows for the determination of a particular solution without having to guess or use trial and error methods. Additionally, the Wronskian method can be used to solve a wide range of ODEs, making it a versatile tool for scientists and mathematicians.

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