Does anybody know how to solve the system of differential equations,

y'(t) = -ay(t) + ax(t)

x'(t) = ay(t) -ax(t)

with initial conditions, y(0)=1 and x(0)=0 where a is a real constant and also with y(t) + x(t) = 1 for all t

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- May 14th 2008, 08:16 AMjohnbarkwithsystem of differential equations
Does anybody know how to solve the system of differential equations,

y'(t) = -ay(t) + ax(t)

x'(t) = ay(t) -ax(t)

with initial conditions, y(0)=1 and x(0)=0 where a is a real constant and also with y(t) + x(t) = 1 for all t - May 14th 2008, 08:40 AMflyingsquirrel
Hi

Let's use the equations we're given : substitute $\displaystyle x(t)=1-y(t) $ in $\displaystyle y'(t)=-ay(t)+ax(t)$ :

$\displaystyle y'(t)=-ay(t)+a(1-y(t))=-2ay(t)+a $

Hence $\displaystyle y$ is given by the ODE $\displaystyle y'(t)+2ay(t)=a$ which you can solve. Once you know $\displaystyle y$, you can use $\displaystyle x(t)=1-y(t)$ to find $\displaystyle x$.