a system of differential equations with initial value problem.

mirov

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Joined
Jul 16, 2018
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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.
 
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.

It's a lovely system, I think.

1) If you are at Differential Equations, and you are online, it is time for you to learn a little LaTex. It will help you communicate an SHOW YOUR work.
2) What have you worked on? Can you expand to a bigger system with 1st derivatives? Maybe not. Give us something to work with in helping you on you way.
 
… I have been trying a lot …
Please show your work.

You don't necessarily need to learn LaTex, but you do need to learn some kind of notation. When you read the forum guidelines, did you see the link for learning how to format math as text?

For example, we show exponents using the caret symbol: x^2
 
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.
 
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