General solution of differential equation system

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SUMMARY

The general solution of the system of differential equations defined by x' = 10x - 12y and y' = 25x - 30y can be derived using the initial conditions x(0) = A and y(0) = B. The solution process involves transforming the equations and finding the characteristic equation, leading to roots r = 0 and r = -20. The general solutions are expressed as x(t) = Ce^{-20t} and y(t) = \frac{5}{2}Ce^{-20t}, where C is a constant determined by initial conditions. The discussion highlights the importance of using distinct constants in the general solution to avoid confusion with initial conditions.

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Homework Statement



Find the general solution of the system of differential equations
[tex]x'=10x - 12y[/tex]
[tex]y'=25x - 30y[/tex]
(where primes indicate derivatives with respect to t) by using the initial conditions
[tex]x(0)=A[/tex]
[tex]y(0)=B[/tex]

Homework Equations



The Attempt at a Solution



[tex]x''=10x' - 12y'[/tex]
[tex]y'=25x-30y[/tex]
[tex]x''=10x'-12(25x-30y)[/tex]
[tex]y=\frac{10x-x'}{12}[/tex]
[tex]x''=10x'-300x+360(\frac{10x-x'}{12})[/tex]
[tex]x''=10x'-300x+300x-30x'[/tex]
[tex]x''+20x'=0[/tex]
[tex]r^2+20r=0[/tex]
[tex]r(r+20)=0[/tex]
[tex]r=0,-20[/tex]
[tex]x(t)=Ae^{-20t} +B[/tex] [tex]\rightarrow[/tex] [tex]x(0)=A[/tex]
[tex]A=A+B[/tex]
[tex]B=0[/tex]
[tex]y=\frac{10x-x'}{12}=\frac{5}{2}Ae^{-20t} \rightarrow [tex]y(0)=B[/tex]<br /> [tex]0=\frac{5}{2}A[/tex]<br /> [tex]A=0[/tex]<br /> <br /> If A=0 and B=0, then my general solutions x(t),y(t)=0. This is clearly not right. What am I doing wrong? Thanks![/tex]
 
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Use C and D in your general solution so they don't get mixed up with the A and B in the initial conditions.
 

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