How to solve coupled second order differential equations2y″−3y′+2z′+3y+z=e2xy″−3y′+z′+2y−z=0

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zalutaloj9a0f · Accepted Answer

Step 1It is visible after some contemplation that the highest derivatives of z and y occur in the same linear combination in both equations, marking the system as a DAE system of index at least one.To better handle the structure, define u=y′+z for this combination and eliminate z from the system.2u′−4y′+3y+u=e2xu′−2y′+2y−u=0Again the terms in the highest derivatives y&#039;,u&#039; occur in the same linear combination. So set v=u−2y=y′+z−2y and eliminate u against v to get an explicit index-1 DAE system.2v′+5y+v=e2xv′−v=0This now can be isolated for the highest order derivatives (or non-derivatives) (y,v&#039;) to get5y=e2x−3vv′=vThis system, which resulted just from easily reversible variable substitutions, now indeed has only one integration constant.So, inserting backwards,v=Cex,y=15e2x−3Cex,u=v+2y=152e2x−Cexz=u−y′=152e2x−Cex−152e2x−3Cex=25Cex

Malia Booth · Accepted Answer

Step 1Define Y(s)=L(y(t)) and Z(s)=L(z(t)) where L denotes the laplace transform.Put a=y(0), b=y′(0), c=z(0)With a little algebra your DE becomes(2s2−3s+3)Y(s)+(2s+1)Z(s)=−as+1s−2+2b+2c(s2−3s+2)Y(s)+(s−1)Z(s)=as−3a+b+cThis can be expressed as a linear system(2s2−3s+32s+1s2−3s+2s−1)(Y(s)Z(s))=(−as+1s−2+2b+2cas−3a+b+c)Can you finish? Keep in mind, once you solve this system via the inverse laplace transform L−1, the constants a,b,c,d will be your arbitrary constants in your general solution.

How to solve coupled second order differential equations

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