Hey guys,
I am new here, and i actually registrated here because i need help with differential equations!
I have the following system of differential equations
y1''= 6y1+30y2
y2''=-5y1-19y2
y1(0)=0
y2(0)=0
y'1(0)=-8
y'2(0)=3
I need to solve that to finde the value y1(3Pi/4)
could you please help me? I have been trying a lot to solve it.
Thanks in advanced.
Here is one method (typically given in any introductory d.e. text):
Given y1''= 6y1+30y2, differentiate two more times
y1'''= 6y1'+ 30y2'
y1''''= 6y1''+ 30y2''
Now since y2''=-5y1-19y2 we can replace y2'', above, with that
y1''''= 6y1+ 30(-5y1- 19y2)= 6y1- 150y1- 570y2= -144y1- 570y2
From the first equation, y1''= 6y1+30y2, 30y2= y1''- 6y1 so 570y2= 19(30y2)= 19y1''- 114y1 and we have
y1''''= -144y1- 19y1''+ 144y1= -19y1'' or
y1''''+ 19y1''= 0.
That is a fourth order linear differential equation with constant coefficients in y1 only. It's characteristic equation is r^4- 19r^2= r^2(r^2- 19)= 0 which has roots r= 0 (a double root), sqrt(19), and -sqrt(19). The general solution to the differential equation is y1= A+ Br+ Ce^{t sqrt(19)}+ De^{-t sqrt(19)}, where A, B, C, and D are undetermined constants.
We also have that 30y2= y1''- 6y1. From y1= A+ Br+ Ce^{t sqrt(19)}+ De^{-t sqrt(19)}, y1'= B+ sqrt(19)C e^{t sqrt(19)}- sqrt(19)D e^{-t sqrt(19)} and
y1''= 19C e^{t sqrt(19)}+ 19D e^{-t sqrt(19)}
30y2= y1''- 6y1= 19C e^{t sqrt(19)}+ 19D e^{-t sqrt(19)}- (A+ Br+ Ce^{t sqrt(19)}+ De^{-t sqrt(19)})= -A- Br+ 18C e^{t sqrt(19)}+ 18D e^{-t sqrt(19)}
y2(t)= -A/30- Br/30+ (9/15)C e^{t sqrt(19)+ (9/15)D e^{-t sqrt(19)
Putting those together
y1= A+ Br+ Ce^{t sqrt(19)}+ De^{-t sqrt(19)}
y2(t)= -A/30- Br/30+ (9/15)C e^{t sqrt(19)+ (9/15)D e^{-t sqrt(19)
To determine A, B, C, and D use the initial conditions.