Help with Second Order Differential Equation with Two Dependent Variables

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Tygra

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Dear Forum,

I am trying to get to grips with a Second Order Differential Equation with Two Dependent Variables, but I am stuck on where to even begin really. I have read to convert the second order to first order equations but I'm struggling still.

So, if I have the equations:

[math]y''a + u'b - c = 0[/math]
[math]u''b - y'b = -q[/math]
Where a, b, c and q are constants. The equations are with respect to z.

I would like to know how exactly I solve for y and u in this scenario?

Many thanks
 
I integrated the first equation and got
[math]\begin{array}{lll} cz&=ay'+bu\\ y'&=\dfrac{u''b+q}{b}=\dfrac{cz-bu}{a}\\ 0&=au''+\dfrac{qa}{b}-cz+bu\\ 0&=\dfrac{b}{a}u +u''+\dfrac{q}{b}+\dfrac{c}{a}z\\ 0&=u''+\alpha u+ \beta z +\gamma \end{array}[/math]
Then I rewrote it as [imath] y''+ ay+bx+c=0 [/imath] such that WA could understand it and found
[math] y(x) = -\dfrac{b}{a}x - \dfrac{c}{a} + k_2 \sin(\sqrt{a} x) + k_1 \cos(\sqrt{a} x). [/math]

This means in your original notation (if I made no substitution errors)
[math] u(z)=\dfrac{c}{b}z-\dfrac{qa^2}{b^2} +k_2\sin\left(z\sqrt{\dfrac{b}{a}}\right)+k_1\cos\left(z\sqrt{\dfrac{b}{a}}\right) [/math]
Finding [imath] y(z) [/imath] requires another integration.
 
Tygra said:
Dear Forum,

I am trying to get to grips with a Second Order Differential Equation with Two Dependent Variables, but I am stuck on where to even begin really. I have read to convert the second order to first order equations but I'm struggling still.

So, if I have the equations:

[math]y''a + u'b - c = 0[/math].....................(1)

[math]u''b - y'b = -q[/math]........................(2)

Where a, b, c and q are constants. The equations are with respect to z.

I would like to know how exactly I solve for y and u in this scenario?

Many thanks
Click to expand...
Another way
Differentiate (1) one more time and add to (2) to eliminate u and continue......
 
Do you study linear algebra? If you understand it, you can define a matrix.

Define.

[imath]Z = \begin{bmatrix} X \\ X' \end{bmatrix} = \begin{bmatrix} y \\ u \\ y' \\ u'\end{bmatrix}[/imath]

This reduce the system to first order.

[imath]Z' = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -b/a \\ 0 & 0 & 1 & 0 \end{bmatrix}\!\!Z + \begin{bmatrix} 0\\ 0 \\ c/a \\ -q/b \end{bmatrix}[/imath]

If you don't undestand to solve the system by matrix, forget this method.
 
Hi guys, thank you for all your responses.

I have looked at all three of your posts and they have helped me out.

However, sorry, in the first equation c should have been y.

I first tried Khansaheb's advice.

So I have:

[math]y''a + u'b - y = 0[/math][math]u''b - y'b = -q[/math]
So I differentiate the first equation giving:

[math]y'''a + u''b - y' = 0[/math]
I then subtract equation 2 from 1, giving:

[math]y'''a + y'b - y = q[/math]
I then convert this to 3 first order differential equations.

[math]x_1 = y[/math][math]x_2 = y_1' = x_1'[/math][math]x_3 = y_2' = x_2'[/math]
Thus, the 3 equations are:

[math]x_1' = x_2[/math][math]x_2' = x_3[/math][math]x_3' = \frac{q}{a} + \frac{b}{a}x_3 + \frac{1}{EI}x_2[/math]
And now I put into a matrix like nasi112 said and use MATLAB to solve it, but it comes out as "INF".

Please bare in mind I am not that experienced with these sort of problems, so I am certain I have gone wrong somewhere.
 
The essential path is to eliminate one ot the two functions and solve the resulting differential equation.

Before I answer, and we continue to speak about different equations due to typos, could you restate the original two equations? Can we write them as
[math]\begin{array}{lll} 0&=y''(x)a -y(x)+ u'(x)b \\ 0&=- y'(x)b +q+u''(x)b \end{array}[/math]and would this be correct?

If so, then we can differentiate the first one, multiply it by [imath] -1 [/imath] and receive
[math]\begin{array}{lll} 0&=-y'''(x)a +y'(x)- u''(x)b \\ 0&=- y'(x)b +q+u''(x)b \end{array}[/math]
Adding them results in
[math] 0=-y'''(x)a +(1-b)y'(x)+q [/math]which can be solved to
[math] y(x) = c_1e^{\alpha x}+c_2e^{-\alpha x}-\dfrac{q}{\beta}x+c_3 [/math]where [imath] \alpha:=\dfrac{\sqrt{1-b}}{\sqrt{a}}\, , \,\beta=1-b. [/imath] I would then use the second equation to solve for [imath] u(x) [/imath]:
[math] u''(x)=y'(x)-\dfrac{q}{\beta}=\alpha c_1e^{\alpha x}-\alpha c_2e^{-\alpha x}-\dfrac{2q}{\beta}+c_3\,. [/math]
Double integration gives you [imath] u(x). [/imath]
 
Hi fresh_42, that is great, thanks.

However, for alpha, in my actual problem I am solving b is greater than 1. So I get an error when I compute alpha.
 
Tygra said:
Hi fresh_42, that is great, thanks.

However, for alpha, in my actual problem I am solving b is greater than 1. So I get an error when I compute alpha.
Click to expand...
In this case change the sign and solve [imath] -ay'''(x)-(1-b)y'(x)+q=0. [/imath] The result is basically the same, just with [imath] b-1 [/imath] instead. [imath] b=1 [/imath] has to be treated separately.
 
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