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Linear_Differential_Equation

1.Solve the following systems by either substitution or elimination:

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

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

Solution:-

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

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

(1)  \(\Rightarrow \) 

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

     \(\frac{\mathrm{d}y}{\mathrm{d} t} = \frac{\mathrm{d}^2 x}{\mathrm{d} t^2}\)          --------------(3)

Plug (3) in (2)

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

     \(\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}\) - x = cos(t)   ----------(4)

A.E.

      \(m^{2} - 1 = 0\)

       m = \(\pm \) 1

      \(x_{c} = C_{1}e^{t} + C_{2}e^{-t}\)

      \(x_{p} = A cos(t) + B sin(t)\)

      \(\frac{\mathrm{d} x_{p}}{\mathrm{d} t} = -A sin(t) + B cos(t)\)

      \(\frac{\mathrm{d}^2 x_{p}}{\mathrm{d} t^2} = -A cos(t) - B sin(t)\)

(4) \(\Rightarrow \)

    - A cos(t) - B sin(t) - A cos(t) - B sin(t) = cos(t)

    - 2A cos(t) - 2B sin(t) = cos(t)

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

     -2B = 0    \(\Rightarrow \)  B = 0

      \(x_{p} = -\frac{1}{2}\) cos(t)

       x = \(x_c + x_p\) 

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

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

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

\(\therefore \)  x(t) = \C_{1}e^{t} + C_{2}e^{-t}\) -\frac{1}{2}\) cos(t)

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


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