Simplifying, two 2nd order linear differential equations

[bsktr]

I'm working on a computer programming project on matlab,
and need an equation in the form of y'' + y' + y = 0 or k or inthe form of y' + y = 0 or k..
everything will be simple for me after that

but I have these two equations

N - ay'' - by' - cy - dx'' - ex' - fx = 0
M - gy'' - hy' - iy - jx'' - kx' - lx = 0 where a,b,c,d,e,f,g,h,i,j,k,l are constants N,M=c sinwt and y(0)=x(0)=y'(0)=x'(0)=0

can you propose me method to be able to arrange these. I didn't deal with diff. eqns . more than 3 years, I may need some help to remember,

if you can help, please do so, I don't have much time :(

 

[lippyka]

First, let's combine the two equations into a single equation. Notice that the terms with y and y'' are identical in both equations. We can subtract one equation from the other to eliminate those terms:N - M = ay'' - by' - cy + dx'' + ex' + fx - gy'' + hy' + iy - jx'' - kx' - lx Now we have an equation of the form: N - M = a(y'' - g) - b(y' - h) - c(y - i) + d(x'' - j) + e(x' - k) + f(x - l)To solve this equation, we need to express the variables y, y'', x, x'', and x' in terms of y''. To do this, we will use the initial conditions given above. Since y(0) = x(0) = y'(0) = x'(0) = 0, we can use the following identities:y = y' = x = x' = 0 y'' = -(a/g)y' - (b/g)y - (c/g) - (d/g)x' - (e/g)x - (f/g)Substituting these expressions into our equation, we obtain:N - M = a(-g - (a/g)y' - (b/g)y - (c/g) - (d/g)x' - (e/g)x - (f/g)) - b(y' - h) - c(y - i) + d(x'' - j) + e(x' - k) + f(x - l)Simplifying the equation, we obtain:N - M = -ag - abh - aci - adk - ael - afl - bh - ci + dj + ek + flThis equation can be written as a differential equation of the form y'' + y' + y = k, where k is a constant. To do this, we need to solve for y'' in terms of the other terms. Rearranging the equation, we obtain:y'' = (N -