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## Homework Statement

Consider the system of coupled second-order differential equations

[tex]u''-(t+1)(u')^2+2uv-u^3=\cos{t}[/tex]

[tex]2v''+(\sin{t})u'v'-6u=2t+3[/tex]

with initial conditions u(0)=1, u'(0)=2, v(0)=3, v'(0)=4. Use the second order Runge-Kutta method with h=0.2 and a=2/3, b=1/3, [tex]\alpha=\beta=3/2[/tex] to find u, u', v, v' at t=0.2

## Homework Equations

For the equation dy/dx=f(x,y)

[tex]y_{n+1}=y_n+ak_1+bk_2[/tex]

[tex]k_1=hf(x_n,y_n)[/tex]

[tex]k_2=hf(x_n+\alpha h,y_n+\beta k_1)[/tex]

## The Attempt at a Solution

I have no background in Differential Equations but some searching around on the net gave me a little bit of insight and I believe I'm supposed to change the two given equations into 4 separate first-order ODE's.

So what I have is as follows:

Rearranging the two 2nd order equations gives

[tex]u''=\cos{t}+(t+1)(u')^2-2uv+u^3 [/tex]

[tex]v''=t+\frac{3}{2}-\frac{1}{2}(\sin{t})u'v'+3u[/tex]

If we now let

[tex]u'=x[/tex]

[tex]v'=y[/tex]

then

[tex]u''=x'[/tex]

[tex]v''=y'[/tex]

and we have

[tex]u'=f(x)=x[/tex]

[tex]v'=m(y)=y[/tex]

[tex]x'=g(t,x,u,v)=\cos{t}+(t+1) (x)^2-2uv+u^3[/tex]

[tex]y'=n(t,x,y,u)=t+\frac{3}{2}-\frac{1}{2}(\sin{t})xy+3u[/tex]

Furthermore

[tex]u(0)=1[/tex]

[tex]u'(0)=x_0=2[/tex]

[tex]v(0)=3[/tex]

[tex]v'(0)=y_0=4[/tex]

so that

[tex]k_{1u}=hf(x_0)

=(0.2)f(2)

=(0.2)(2)

=0.4[/tex]

[tex]k_{1v}=hm(y_0)

=(0.2)m(4)

=(0.2)(4)

=0.8[/tex]

[tex]k_{1x}=hg(t,x_0,u,v)

=(0.2)g(0,2,1,3)

=(0.2)[1+(2)^2-2(1)(3)+(1)^3]

=0[/tex]

[tex]k_{1y}=hn(t,x_0,y_0,u)

=(0.2)n(0,2,4,1)

=(0.2)[(0)+\frac{3}{2}-(0)+3(1)]

=0.9[/tex]

*Assuming*all of the above is correct then, well, what now? Do I say, e.g.

[tex]k_{2u}=hf(x_0+\alpha h)[/tex]

[tex]k_{2v}=hm(y_0+\alpha h)[/tex]

or is it supposed to be

[tex]k_{2u}=hf(x_0+\alpha k_{1u})[/tex]

[tex]k_{2v}=hm(y_0+\alpha k_{1v})[/tex]

?

And what about [tex]k_{2x}[/tex] and [tex]k_{2y}[/tex], each of which has four variables?

Thanks for your help!

phyz

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