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.

ARNAB2012
  • 1
  • how this question can be off-topic ? I mean this is the whole mathematics question . The topic belongs to "homogeneous differential equation of second order" , Wronskian theorem , linearly independent solutions of equation . – ARNAB2012 Apr 06 '16 at 06:01
  • Are you sure the differential equation is not $$x''(t) \color{red}{-} 5x'(t) + 4x(t) = 0$$ instead? – Did Apr 15 '16 at 06:19

2 Answers2

0

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.

Archisman Panigrahi
  • 2,156
  • 2
  • 19
  • 30
-1

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 $$\left(\matrix{x\\y}\right)' = \left(\matrix{-4 & 1 \\ 0 & -1}\right) \left(\matrix{x\\y}\right)$$

Now rewrite the given system of equations to the same form, too: $$\left(\matrix{x\\y}\right)' = \left(\matrix{3 & 2 \\ 1 & 2}\right) \left(\matrix{x\\y}\right)$$ 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 $x_{1/2}(t)$

Ilja
  • 209
  • Can you Please Explain ! I mean how The system $y = x' + 4x$ and $y' + y = 0$ is equivalent to the second-order-equation ? – ARNAB2012 Apr 06 '16 at 07:00
  • Well, just insert the $y$ from the first equation into the second, and you get the desired second-order-equation for $x$ – Ilja Apr 06 '16 at 09:32
  • That is understood. But how do I calculate general solution for equation 1 in terms of x1 and x2 ??? – ARNAB2012 Apr 07 '16 at 05:42