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MATHMAA

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 = $$x_c + x_p$$

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

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

$$y = \frac{\mathrm{d} x}{\mathrm{d} t} - x -t$$

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

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