Solve the initial value problem x1 x1 x2 exp2t x2 x1 3x

 Solve the initial value problem   x_1\' =  x_1 + x_2 + exp(2t)  x_2\' =  -x_1 + 3x_2   where x_1(1)=0, x_2(1)=0.  

Solution

x1\'-x2\'=2x1-2x2=2(x1-x2)=(x1-x2)\'

So, Integrating gives

x1-x2=Ae^{2t}

x1=x2+Ae^{2t}

x2\'=-x2-Ae^{2t}+3x2

x2\'-2x2=-Ae^{2t}

INtegrating factor is exp(-2t)

(x2\'-2x2)exp(-2t)=-A

(x2 exp(-2t))\'=-A

x2 exp(-2t)=-At+B

x2=exp(2t)(-At+B)

x1=x2+Ae^{2t}=exp(2t)(-At+A+B)

x1(1)=0=-A+A+B

So, B=0

x2(1)=A=0

SO trivial solution