Solve the simultaneous differential equations x′ + x + y′/2 = 0 x(0) = y(0) = 0 x′/2 + y′ + y = 0

Question

Solve the simultaneous differential equations

[latex]\begin{aligned} & x^{\prime}+x+\frac{y^{\prime}}{2}=1 \quad x(0)=y(0)=0 \ & \frac{x^{\prime}}{2}+y^{\prime}+y=0 \end{aligned}[/latex]

Accepted Answer

Take the Laplace transforms of both equations:

[latex]\begin{aligned} & s X(s)-x(0)+X(s)+\frac{s Y(s)-y(0)}{2}=\frac{1}{s} \ & \frac{s X(s)-x(0)}{2}+s Y(s)-y(0)+Y(s)=0 \end{aligned}[/latex]

These are rearranged to give

[latex]\begin{aligned} & (s+1) X(s)+\frac{s Y(s)}{2}=\frac{1}{s} \ & \frac{s X(s)}{2}+(s+1) Y(s)=0 \end{aligned}[/latex]

These simultaneous algebraic equations need to be solved for X(s) and Y(s). By Cramer's rule (see Section 8.7.2)

[latex]Y(s)=\frac{\left|\begin{array}{cc} s+1 & 1 / s \ s / 2 & 0 \end{array} ight|}{\left|\begin{array}{cc} s+1 & s / 2 \ s / 2 & s+1 \end{array} ight|}=\frac{-1 / 2}{(s+1)^2-s^2 / 4} \ \begin{aligned} & =-\frac{1}{2}\left\{\frac{1}{3 s^2 / 4+2 s+1} ight\} \ & =-\frac{1}{2}\left\{\frac{4}{3 s^2+8 s+4} ight\} \ & =-\frac{1}{2}\left\{\frac{4}{(3 s+2)(s+2)} ight\} \end{aligned}[/latex]

Using partial fractions we find

[latex]\begin{aligned} Y(s) & =-\frac{1}{2}\left\{\frac{3}{3 s+2}-\frac{1}{s+2} ight\} \ & =-\frac{1}{2}\left\{\frac{1}{s+\frac{2}{3}}-\frac{1}{s+2} ight\} \end{aligned}[/latex]

and hence

[latex]y(t)=\frac{1}{2}\left(\mathrm{e}^{-2 t}-\mathrm{e}^{-2 t / 3} ight)[/latex]

Similarly,

[latex]X(s)=\frac{4(s+1)}{s(3 s+2)(s+2)}=\frac{1}{s}-\frac{1}{2\left(s+\frac{2}{3} ight)}-\frac{1}{2(s+2)}[/latex]

and so

[latex]x(t)=1-\frac{1}{2}\left(\mathrm{e}^{-2 t / 3}+\mathrm{e}^{-2 t} ight)[/latex]

Question 21.31: Solve the simultaneous differential equations x′ + x + y′/2......

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