online user counteronline user counter

MATHMAA

Only Search Your Site

Probability Introduction :

 b) \(\frac{\mathrm{d} x}{\mathrm{d} t} = x + y +t\)

      \(\frac{\mathrm{d} y}{\mathrm{d} t} = - x + y - 1\)

Solution:-

      \(\frac{\mathrm{d} x}{\mathrm{d} t} = x + y +t\)     --------------(1)

      \(\frac{\mathrm{d} y}{\mathrm{d} t} = - x + y - 1\)  --------------(2)

       y = \(\frac{\mathrm{d} x}{\mathrm{d} t}\) - x - t

      \(\frac{\mathrm{d} y}{\mathrm{d} t}\) = \(\frac{\mathrm{d}^2 x}{\mathrm{d}  t^2}\) - \(\frac{\mathrm{d} x}{\mathrm{d} t}\) - 1

      \(\frac{\mathrm{d}^2 x}{\mathrm{d}  t^2}\) - \(\frac{\mathrm{d} x}{\mathrm{d} t}\) - 1 = -x + \(\frac{\mathrm{d} x}{\mathrm{d} t}\) - x - t - 1

     \(\frac{\mathrm{d}^2 x}{\mathrm{d}  t^2}\) - \(\frac{2\mathrm{d} x}{\mathrm{d} t}\) + 2x = t

A.E

    \(m^2 - 2m + 2 = 0\)

     \(m = 1 \pm i \)

      \(x_c = e^t (C_1 cos(t)+ C_2 sin(t))\)

      \(x_p =At + B\)

       \(x'_p = A\)

       \(x''_p = 0\)

    0 - 2A + 2At + 2B = -t

    2A = -1   \(\Rightarrow \)  A = -\(\frac{1}{2}\)  

    -2A + 2B = 0

      B = A

      B = -\(\frac{1}{2}\)

     \(x_p = -\frac{1}{2}t - \frac{1}{2}\)

     x = \(x_c + x_p\)

    \(x(t)=e^t(C_1 cos(t)+ C_2 sin(t)-\frac{1}{2}t - \frac{1}{2} )\)

\(\frac{\mathrm{d} x}{\mathrm{d} t}  = e^t(C_1 cos(t) + C_2 sin(t))+ e^t(-C_1 sin(t) + C_2 cos(t)) - \frac{1}{2}\)

\(y = \frac{\mathrm{d} x}{\mathrm{d} t} - x -t\)

\(y  = e^t(-C_1 sin(t) + C_2 cos(t)) - \frac{1}{2}t\)

  

<<BACK<<                                                                            >> NEXT >>


SHARE YOU ENORMOUS EFFORT AND SMART EXAMPLES HERE

!! NEED MORE HELP !!

SBI! Case Studies