Recent content by Arnab Patra

calculating eigenvalues are 4 and 1 . Eigen vectors are (2,1) and (-1,1) and $$\left(\matrix{x_1(t) \\ x_2(t)}\right)=c_1e^{4t} \left(\matrix{2 \\ 1 }\right) +c_2e^{t} \left(\matrix{-1 \\ 1 } \right)$$ or $$\left(\matrix{x_1(t) \\ x_2(t)}\right)=\left(\matrix{2 & -1\\ 1 & 1 }\right)...

sorry guys . i was wrong about the question .. i found the question and i am attaching the question's image

by considering $$x''(t)+5x'(t)+4x(t)=0$$ equation as $$r^2+5r+4=0$$ i get $$r=-4,-1$$ i.e. $$x'(t)=-4,-1$$ which are the two roots of the said equation. from this solution i found the general equation of the said equation i.e. $$x(t) = c_1e^{-4t} + c_2e^{-t}$$ . And you are asking to solve...

sorry my mistake .. $$x _1(t) = 3/2 $$ and $$x _2(t) = -1/2 $$ .

The question is correct . X1 = -4 and x2 = -1 . And now what ?

Suppose ##x_1(t)## and ##x_2(t)## are two linearly independent solutions of the equations: ##x'_1(t) = 3x_1(t) + 2x_2(t)## and ##x'_2(t) = x_1(t) + 2x_2(t)## where ##x'_1(t)\text{ and }x'_2(t)## denote the first derivative of functions ##x_1(t)## and ##x_2(t)## respectively with respect to...