题目来源
常微分方程(第四版) (王高雄,周之铭,朱思铭,王寿松) 高等教育出版社
书中习题4.1
对应知识
3.已知齐次线性微分方程的基本解组 x 1 , x 2 x_1,x_2 x1,x2，求下列方程对应的非齐次线性微分方程的通解
三道题都是常数变易法
(1)
x ′ ′ − x = cos t , x 1 = e t , x 2 = e − t x''-x=\cos t,x_1=e^t,x_2=e^{-t} x′′−x=cost,x1=et,x2=e−t
解
x ( t ) = c 1 ( t ) e t + c 2 ( t ) e − t x(t)=c_1(t)e^t+c_2(t)e^{-t} x(t)=c1(t)et+c2(t)e−t
{ c 1 ′ ( t ) e t + c 2 ′ ( t ) e − t = 0 c 1 ′ ( t ) e t − c 2 ′ ( t ) e − t = cos t \begin{cases} c'_1(t)e^t+c'_2(t)e^{-t}=0\\ c'_1(t)e^t-c'_2(t)e^{-t}=\cos t \end{cases} {c1′(t)et+c2′(t)e−t=0c1′(t)et−c2′(t)e−t=cost
解得
{ c 1 ′ ( t ) = 1 2 e − t cos t c 2 ′ ( t ) = − 1 2 e t cos t \begin{cases} c'_1(t)=\frac{1}{2}e^{-t}\cos t\\ c'_2(t)=-\frac{1}{2}e^{t}\cos t \end{cases} {c1′(t)=21e−tcostc2′(t)=−21etcost
积分得
{ c 1 ( t ) = − 1 4 e t ( cos t − sin t ) + c 1 c 2 ( t ) = − 1 4 e t ( cos t + sin t ) + c 2 \begin{cases} c_1(t)=-\frac{1}{4}e^{t}(\cos t-\sin t)+c_1\\ c_2(t)=-\frac{1}{4}e^{t}(\cos t+\sin t)+c_2 \end{cases} {c1(t)=−41et(cost−sint)+c1c2(t)=−41et(cost+sint)+c2
故所求通解为
x ( t ) = c 1 e t + c 2 e − t − 1 2 cos t x(t)=c_1e^t+c_2e^{-t}-\frac{1}{2}\cos t x(t)=c1et+c2e−t−21cost
(2)
x ′ ′ + t 1 − t x ′ − 1 1 − t x = t − 1 , x 1 = t , x 2 = e t x''+\frac{t}{1-t}x'-\frac{1}{1-t}x=t-1,x_1=t,x_2=e^{t} x′′+1−ttx′−1−t1x=t−1,x1=t,x2=et
解
x ( t ) = c 1 ( t ) t + c 2 ( t ) e t x(t)=c_1(t)t+c_2(t)e^{t} x(t)=c1(t)t+c2(t)et
{ c 1 ′ ( t ) t + c 2 ′ ( t ) e t = 0 c 1 ′ ( t ) + c 2 ′ ( t ) e t = t − 1 \begin{cases} c'_1(t)t+c'_2(t)e^{t}=0\\ c'_1(t)+c'_2(t)e^{t}=t-1 \end{cases} {c1′(t)t+c2′(t)et=0c1′(t)+c2′(t)et=t−1
解得
{ c 1 ′ ( t ) = − 1 c 2 ′ ( t ) = t e t \begin{cases} c'_1(t)=-1\\ c'_2(t)=te^{t} \end{cases} {c1′(t)=−1c2′(t)=tet
积分得
{ c 1 ( t ) = − t + c 1 c 2 ( t ) = − e − t ( t + 1 ) + c 2 \begin{cases} c_1(t)=-t+c_1\\ c_2(t)=-e^{-t}(t+1)+c_2 \end{cases} {c1(t)=−t+c1c2(t)=−e−t(t+1)+c2
通解
x ( t ) = c 1 t + c 2 e t − ( t 2 + t + 1 ) x(t)=c_1t+c_2e^t-(t^2+t+1) x(t)=c1t+c2et−(t2+t+1)
(3)
x ′ ′ + 4 x = t sin 2 t , x 1 = cos 2 t , x 2 = sin 2 t x''+4x=t\sin2t,x_1=\cos2t,x_2=\sin2t x′′+4x=tsin2t,x1=cos2t,x2=sin2t
解
x ( t ) = c 1 ( t ) cos 2 t + c 2 ( t ) sin 2 t x(t)=c_1(t)\cos2t+c_2(t)\sin2t x(t)=c1(t)cos2t+c2(t)sin2t
{ c 1 ′ ( t ) cos 2 t + c 2 ′ ( t ) sin 2 t = 0 − 2 c 1 ′ ( t ) sin 2 t + 2 c 2 ′ ( t ) cos 2 t = t sin 2 t \begin{cases} c'_1(t)\cos2t+c'_2(t)\sin2t=0\\ -2c'_1(t)\sin2t+2c'_2(t)\cos2t=t\sin2t \end{cases} {c1′(t)cos2t+c2′(t)sin2t=0−2c1′(t)sin2t+2c2′(t)cos2t=tsin2t
解得
{ c 1 ′ ( t ) = 1 4 t ( cos 4 t − 1 ) c 2 ′ ( t ) = 1 4 t sin 4 t \begin{cases} c'_1(t)=\frac{1}{4}t(\cos4t-1)\\ c'_2(t)=\frac{1}{4}t\sin4t \end{cases} {c1′(t)=41t(cos4t−1)c2′(t)=41tsin4t
积分得
{ c 1 ( t ) = 1 64 cos 4 t + 1 16 t sin 4 t − 1 8 t 2 + γ 1 c 2 ( t ) = 1 64 sin 4 t − 1 16 t cos 4 t + γ 2 \begin{cases} c_1(t)=\frac{1}{64}\cos4t+\frac{1}{16}t\sin4t-\frac{1}{8}t^2+\gamma_1\\ c_2(t)=\frac{1}{64}\sin4t-\frac{1}{16}t\cos4t+\gamma_2 \end{cases} {c1(t)=641cos4t+161tsin4t−81t2+γ1c2(t)=641sin4t−161tcos4t+γ2
通解
x ( t ) = c 1 cos 2 t + c 2 sin 2 t − 1 8 t 2 cos 2 t + 1 16 t sin 2 t x(t)=c_1\cos2t+c_2\sin2t-\frac{1}{8}t^2\cos2t+\frac{1}{16}t\sin2t x(t)=c1cos2t+c2sin2t−81t2cos2t+161tsin2t