习题7-10

1．求下列微分方程组的通解： （1）$\left\{\begin{array}{l}\frac{\mathrm{d} y}{\mathrm{~d} x}=z, \\ \frac{\mathrm{~d} z}{\mathrm{~d} x}=y ;\end{array}\right.$ （2）$\left\{\begin{array}{l}\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}=y, \\ \frac{\mathrm{~d}^{2} y}{\mathrm{~d} t^{2}}=x ;\end{array}\right.$ （3）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}+\frac{\mathrm{d} y}{\mathrm{~d} t}=-x+y+3 \\ \frac{\mathrm{~d} x}{\mathrm{~d} t}-\frac{\mathrm{d} y}{\mathrm{~d} t}=x+y-3\end{array}\right.$ ， （4）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}+5 x+y=\mathrm{e}^{t} \\ \frac{\mathrm{~d} y}{\mathrm{~d} t}-x-3 y=\mathrm{e}^{2 t}\end{array}\right.$ ； （5）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x+\frac{\mathrm{d} y}{\mathrm{~d} t}+y=t, \\ 5 x+\frac{\mathrm{d} y}{\mathrm{~d} t}+3 y=t^{2} ;\end{array}\right.$ （6）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}-3 x+2 \frac{\mathrm{~d} y}{\mathrm{~d} t}+4 y=2 \sin t \\ 2 \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x+\frac{\mathrm{d} y}{\mathrm{~d} t}-y=\cos t .\end{array}\right.$

2．求下列微分方程组满足所给初值条件的特解： （1）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}=y,\left.x\right|_{t=0}=0, \\ \frac{\mathrm{~d} y}{\mathrm{~d} t}=-x,\left.y\right|_{t=0}=1 ;\end{array}\right.$ （2）$\left\{\begin{array}{l}\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}}+2 \frac{\mathrm{~d} y}{\mathrm{~d} t}-x=0,\left.x\right|_{t=0}=1, \\ \frac{\mathrm{~d} x}{\mathrm{~d} t}+y=0,\left.y\right|_{t=0}=0 ;\end{array}\right.$ （3）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}+3 x-y=0,\left.x\right|_{t=0}=1, \\ \frac{\mathrm{~d} y}{\mathrm{~d} t}-8 x+y=0,\left.y\right|_{t=0}=4 ;\end{array}\right.$ （4）$\left\{\begin{array}{l}2 \frac{\mathrm{~d} x}{\mathrm{~d} t}-4 x+\frac{\mathrm{d} y}{\mathrm{~d} t}-y=\mathrm{e}^{t},\left.x\right|_{t=0}=\frac{3}{2}, \\ \frac{\mathrm{~d} x}{\mathrm{~d} t}+3 x+y=0,\left.y\right|_{t=0}=0 ;\end{array}\right.$ （5）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}+2 x-\frac{\mathrm{d} y}{\mathrm{~d} t}=10 \cos t,\left.x\right|_{t=0}=2, \\ \frac{\mathrm{~d} x}{\mathrm{~d} t}+\frac{\mathrm{d} y}{\mathrm{~d} t}+2 y=4 \mathrm{e}^{-2 t},\left.y\right|_{t=0}=0 ;\end{array}\right.$ （6）$\left\{\begin{array}{l}\frac{\mathrm{d} x}{\mathrm{~d} t}-x+\frac{\mathrm{d} y}{\mathrm{~d} t}+3 y=\mathrm{e}^{-t}-1,\left.x\right|_{t=0}=\frac{48}{49}, \\ \frac{\mathrm{~d} x}{\mathrm{~d} t}+2 x+\frac{\mathrm{d} y}{\mathrm{~d} t}+y=\mathrm{e}^{2 t}+t,\left.y\right|_{t=0}=\frac{95}{98} .\end{array}\right.$