find general equation of $x''(t) + 5x'(t) + 4x(t) = 0$ 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 $t$. 

Find the general solution of

$x''(t) + 5x'(t) + 4x(t) = 0$ 

in terms of $x_1(t)$ and $x_2(t)$.



The general solution of the equation

$x''(t) + 5x'(t) + 4x(t) = 0$................................(1)

is 

$x(t) = c _1 e^{-4t} + c _2e^{-t}$................................(2)

Now if i want to express equation (2) in term of $x_1(t)$ and $x_2(t)$ , what exactly i have to do ?



This question was asked on Indian Statistical Institute exam sheet some years ago.
Both $x_1(t)=e^{-4t}$ and $x_2(t)=e^{-t}$ are linearly independent solutions of the given differential equation.

Thus, the complete general solution is, 
$x(t) = c_1 x_1(t)+c_2x_2(t)=c_1e^{-4t}+c_2e^{-t}$, where $c_1$ and $c_2$ are arbitrary constants.
Consider $a^2 + 5a + 4 = (a+1)(a+4)$
Now "substitute" the variable $a$ by "taking the derivative":
The system $y = x' + 4x$ and $y' + y = 0$ is equivalent to your second-order-equation. This is the important result that you can check, the thoughts above are just a help for guessing correctly.  

You can rewrite the system as $x' = -4x+y$ and $y' = -y$ or $$\left(\matrix{x\\y}\right)' = \left(\matrix{-4 &amp; 1 \\ 0 &amp; -1}\right) \left(\matrix{x\\y}\right)$$

Now rewrite the given system of equations to the same form, too: $$\left(\matrix{x\\y}\right)' = \left(\matrix{3 &amp; 2 \\ 1 &amp; 2}\right) \left(\matrix{x\\y}\right)$$
Now find the transformation between these matrices and you get a basis for the vector space of the solutions of the first system as a linear combination of the solutions of the second, which are the given $x_{1/2}(t)$