Help with Second Order Differential Equation with Two Dependent Variables

[Tygra]

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

 

[fresh_42]

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.

 

[khansaheb]

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

Another way
Differentiate (1) one more time and add to (2) to eliminate u and continue......