本方法只有一个公式，和算子法比减少记忆负担，和待定系数法相比减少计算量。 二阶常系数非齐次线性微分方程： y ′ ′ + p y ′ + q y = f ( x ) ( 1 ) y^{\prime \prime}+p y^{\prime}+q y=f(x)(1) y′′+py′+qy=f(x)(1) 其对应的齐次微分方程为： y ′ ′ + p y ′ + q y = 0 ( 2 ) y^{\prime \prime}+p y^{\prime}+q y=0(2) y′′+py′+qy=0(2) 设 y 1 ( x ) , y 2 ( x ) y_1(x), y_2(x) y1​(x),y2​(x)是方程(2)有两个线性无关的特解，则方程(1)有特解: y ∗ = y 1 ∫ f ( x ) y 2 ( y 1 y 2 ) ′ d x + y 2 ∫ f ( x ) y 1 ( y 2 y 1 ) ′ d x y^*=y_1 \int \frac{f(x)}{y_2\left(\frac{y_1}{y_2}\right)^{\prime}} \mathrm{d} x+y_2 \int \frac{f(x)}{y_1\left(\frac{y_2}{y_1}\right)^{\prime}} \mathrm{d} x y∗=y1​∫y2​(y2​y1​​)′f(x)​dx+y2​∫y1​(y1​y2​​)′f(x)​dx 与一阶线性微分方程求解公式类似，这里不定积分不用+C

首先解出两个齐次特解： r 2 + 2 r + 1 = 0 ⇒ r 1 , 2 = − 1 ⇒ y 1 = x e − x , y 2 = e − x r^2+2 r+1=0 \Rightarrow r_{1,2}=-1 \Rightarrow y_1=x e^{-x}, y_2=e^{-x} r2+2r+1=0⇒r1,2​=−1⇒y1​=xe−x,y2​=e−x 再代公式解出非齐次特解： y ∗ = x e − x ∫ ( 3 x + 2 ) e − x e − x ( x ) ′ d x + e − x ∫ ( 3 x + 2 ) e − x x e − x ( 1 x ) ′ d x = x e − x ∫ ( 3 x + 2 ) d x − e − x ∫ ( 3 x 2 + 2 x ) d x = x e − x ( 3 2 x 2 + 2 x ) − e − x ( x 3 + x 2 ) = 1 2 x 3 e − x + x 2 e − x \begin{aligned} & y^*=x e^{-x} \int \frac{(3 x+2) e^{-x}}{e^{-x}(x)^{\prime}} \mathrm{d} x+e^{-x} \int \frac{(3 x+2) e^{-x}}{x e^{-x}\left(\frac{1}{x}\right)^{\prime}} \mathrm{d} x \\ & =x e^{-x} \int(3 x+2) \mathrm{d} x-e^{-x} \int\left(3 x^2+2 x\right) \mathrm{d} x \\ & =x e^{-x}\left(\frac{3}{2} x^2+2 x\right)-e^{-x}\left(x^3+x^2\right) \\ & =\frac{1}{2} x^3 e^{-x}+x^2 e^{-x} \end{aligned} ​y∗=xe−x∫e−x(x)′(3x+2)e−x​dx+e−x∫xe−x(x1​)′(3x+2)e−x​dx=xe−x∫(3x+2)dx−e−x∫(3x2+2x)dx=xe−x(23​x2+2x)−e−x(x3+x2)=21​x3e−x+x2e−x​

首先解出两个齐次特解： r 2 − 4 r + 3 = 0 ⟹ r 1 = 1 , r 2 = 3 ⟹ y 1 = e x , y 2 = e 3 x r^2-4r+3=0\Longrightarrow r_1=1,r_2=3\Longrightarrow y_1=\mathrm{e}^x,y_2=\mathrm{e}^{3x} r2−4r+3=0⟹r1​=1,r2​=3⟹y1​=ex,y2​=e3x 再代公式解出非齐次特解： y ∗ = e x ∫ 2 e 2 x e 3 x ( e x e 3 x ) ′ d x + e 3 x ∫ 2 e 2 x e x ( e 3 x e x ) ′ d x = e x ∫ 2 e 2 x − 2 e − 2 x ⋅ e 3 x   d x + e 3 x ∫ 2 e 2 x 2 e x ⋅ e 2 x   d x = − e x ∫ e x   d x + e 3 x ∫ e − x   d x = − 2 e 2 x \begin{aligned} y^* & =\mathrm{e}^x \int \frac{2 \mathrm{e}^{2 x}}{\mathrm{e}^{3 x}\left(\frac{\mathrm{e}^x}{\mathrm{e}^{3 x}}\right)^{\prime}} \mathrm{d} x+\mathrm{e}^{3 x} \int \frac{2 \mathrm{e}^{2 x}}{\mathrm{e}^x\left(\frac{\mathrm{e}^{3 x}}{\mathrm{e}^x}\right)^{\prime}} \mathrm{d} x \\ & =\mathrm{e}^x \int \frac{2 \mathrm{e}^{2 x}}{-2 \mathrm{e}^{-2 x} \cdot \mathrm{e}^{3 x}} \mathrm{~d} x+\mathrm{e}^{3 x} \int \frac{2 \mathrm{e}^{2 x}}{2 \mathrm{e}^x \cdot \mathrm{e}^{2 x}} \mathrm{~d} x \\ & =-\mathrm{e}^x \int \mathrm{e}^x \mathrm{~d} x+\mathrm{e}^{3 x} \int \mathrm{e}^{-x} \mathrm{~d} x \\ & =-2 \mathrm{e}^{2 x} \end{aligned} y∗​=ex∫e3x(e3xex​)′2e2x​dx+e3x∫ex(exe3x​)′2e2x​dx=ex∫−2e−2x⋅e3x2e2x​ dx+e3x∫2ex⋅e2x2e2x​ dx=−ex∫ex dx+e3x∫e−x dx=−2e2x​ 因此该方程的通解为： y = c 1 e 3 x + c 2 e x − 2 e 2 x y=c_1 \mathrm{e}^{3 x}+c_2 \mathrm{e}^x-2 \mathrm{e}^{2 x} y=c1​e3x+c2​ex−2e2x

首先解出两个齐次特解： r 2 − 3 r + 2 = 0 ⟹ r 1 = 1 , r 2 = 2 ⟹ y 1 = e x , y 2 = e 2 x r^2-3r+2=0\Longrightarrow r_1=1,r_2=2\Longrightarrow y_1=\mathrm{e}^x,y_2=\mathrm{e}^{2x} r2−3r+2=0⟹r1​=1,r2​=2⟹y1​=ex,y2​=e2x

再代公式解出非齐次特解： y ∗ = e x ∫ 2 x e x e 2 x ( e x e 2 x ) ′ d x + e 2 x ∫ 2 x e x e x ( e 2 x e x ) ′ d x = e x ∫ 2 x e x − e x   d x + e 2 x ∫ 2 x e x e 2 x   d x = − e x ∫ 2 x   d x + 2 e 2 x ∫ x e − x   d x = − x 2 e x − 2 e 2 x ( x e − x + e − x ) = − x ( x + 2 ) e x + 2 e x \begin{aligned} y^* & =\mathrm{e}^x \int \frac{2 x \mathrm{e}^x}{\mathrm{e}^{2 x}\left(\frac{\mathrm{e}^x}{\mathrm{e}^{2 x}}\right)^{\prime}} \mathrm{d} x+\mathrm{e}^{2 x} \int \frac{2 x \mathrm{e}^x}{\mathrm{e}^x\left(\frac{\mathrm{e}^{2 x}}{\mathrm{e}^x}\right)^{\prime}} \mathrm{d} x \\ & =\mathrm{e}^x \int \frac{2 x \mathrm{e}^x}{-\mathrm{e}^x} \mathrm{~d} x+\mathrm{e}^{2 x} \int \frac{2 x \mathrm{e}^x}{\mathrm{e}^{2 x}} \mathrm{~d} x \\ & =-\mathrm{e}^x \int 2 x \mathrm{~d} x+2 \mathrm{e}^{2 x} \int x \mathrm{e}^{-x} \mathrm{~d} x \\ & =-x^2 \mathrm{e}^x-2 \mathrm{e}^{2 x}\left(x \mathrm{e}^{-x}+\mathrm{e}^{-x}\right) \\ & =-x(x+2) \mathrm{e}^x+2 \mathrm{e}^x \end{aligned} y∗​=ex∫e2x(e2xex​)′2xex​dx+e2x∫ex(exe2x​)′2xex​dx=ex∫−ex2xex​ dx+e2x∫e2x2xex​ dx=−ex∫2x dx+2e2x∫xe−x dx=−x2ex−2e2x(xe−x+e−x)=−x(x+2)ex+2ex​ 因此该方程的通解为： y = c 1 e x + c 2 e 2 x − x ( x + 2 ) e x y=c_1 \mathrm{e}^x+c_2 \mathrm{e}^{2 x}-x(x+2) \mathrm{e}^x y=c1​ex+c2​e2x−x(x+2)ex

首先解出两个齐次特解： r 2 + 1 = 0 ⟹ r 1 , 2 = ± i ⟹ y 1 = cos ⁡ x , y 2 = sin ⁡ x r^2+1=0\Longrightarrow r_{1,2}=\pm i\Longrightarrow y_1=\cos x,y_2=\sin x r2+1=0⟹r1,2​=±i⟹y1​=cosx,y2​=sinx 再代公式解出非齐次特解： y ∗ = cos ⁡ x ∫ x cos ⁡ 2 x sin ⁡ x ⋅ ( cos ⁡ x sin ⁡ x ) ′ d x + sin ⁡ x ∫ x cos ⁡ 2 x cos ⁡ x ⋅ ( sin ⁡ x cos ⁡ x ) ′ d x = − cos ⁡ x ∫ x sin ⁡ x cos ⁡ 2 x   d x + sin ⁡ x ∫ x cos ⁡ x cos ⁡ 2 x   d x = − 1 2 cos ⁡ x ∫ x ( sin ⁡ 3 x − sin ⁡ x ) d x + 1 2 sin ⁡ x ∫ x ( cos ⁡ 3 x + cos ⁡ x ) d x = − 1 2 cos ⁡ x ⋅ ( − 1 3 x cos ⁡ 3 x + 1 9 sin ⁡ 3 x + x cos ⁡ x − sin ⁡ x ) + 1 2 sin ⁡ x ⋅ ( 1 3 x sin ⁡ 3 x + 1 9 cos ⁡ 3 x + x sin ⁡ x + cos ⁡ x ) = 1 6 x cos ⁡ x cos ⁡ 3 x − 1 18 cos ⁡ x sin ⁡ 3 x − 1 2 x cos ⁡ 2 x + 1 2 sin ⁡ x cos ⁡ x + 1 6 x sin ⁡ x sin ⁡ 3 x + 1 18 sin ⁡ x cos ⁡ 3 x + 1 2 x sin ⁡ 2 x + 1 2 sin ⁡ x cos ⁡ x = 1 6 x cos ⁡ 2 x − 1 18 sin ⁡ 2 x − 1 2 x cos ⁡ 2 x + 1 2 sin ⁡ 2 x = − 1 3 x cos ⁡ 2 x + 4 9 sin ⁡ 2 x \begin{aligned} y^* & =\cos x \int \frac{x \cos 2 x}{\sin x \cdot\left(\frac{\cos x}{\sin x}\right)^{\prime}} \mathrm{d} x+\sin x \int \frac{x \cos 2 x}{\cos x \cdot\left(\frac{\sin x}{\cos x}\right)^{\prime}} \mathrm{d} x \\ & =-\cos x \int x \sin x \cos 2 x \mathrm{~d} x+\sin x \int x \cos x \cos 2 x \mathrm{~d} x \\ & =-\frac{1}{2} \cos x \int x(\sin 3 x-\sin x) \mathrm{d} x+\frac{1}{2} \sin x \int x(\cos 3 x+\cos x) \mathrm{d} x \\ & =-\frac{1}{2} \cos x \cdot\left(-\frac{1}{3} x \cos 3 x+\frac{1}{9} \sin 3 x+x \cos x-\sin x\right)+\frac{1}{2} \sin x \cdot\left(\frac{1}{3} x \sin 3 x+\frac{1}{9} \cos 3 x+x \sin x+\cos x\right) \\ & =\frac{1}{6} x \cos x \cos 3 x-\frac{1}{18} \cos x \sin 3 x-\frac{1}{2} x \cos ^2 x+\frac{1}{2} \sin x \cos x+\frac{1}{6} x \sin x \sin 3 x+\frac{1}{18} \sin x \cos 3 x+\frac{1}{2} x \sin ^2 x+\frac{1}{2} \sin x \cos x \\ & =\frac{1}{6} x \cos 2 x-\frac{1}{18} \sin 2 x-\frac{1}{2} x \cos 2 x+\frac{1}{2} \sin 2 x \\ & =-\frac{1}{3} x \cos 2 x+\frac{4}{9} \sin 2 x \end{aligned} y∗​=cosx∫sinx⋅(sinxcosx​)′xcos2x​dx+sinx∫cosx⋅(cosxsinx​)′xcos2x​dx=−cosx∫xsinxcos2x dx+sinx∫xcosxcos2x dx=−21​cosx∫x(sin3x−sinx)dx+21​sinx∫x(cos3x+cosx)dx=−21​cosx⋅(−31​xcos3x+91​sin3x+xcosx−sinx)+21​sinx⋅(31​xsin3x+91​cos3x+xsinx+cosx)=61​xcosxcos3x−181​cosxsin3x−21​xcos2x+21​sinxcosx+61​xsinxsin3x+181​sinxcos3x+21​xsin2x+21​sinxcosx=61​xcos2x−181​sin2x−21​xcos2x+21​sin2x=−31​xcos2x+94​sin2x​ 看起来计算量很大，不过很多同类项可以快速合并，所以实际上仍然要比待定系数法计算量小，我们可以对比一下待定系数法的计算量。注意到 2 i 2 i 2i不是特征方程的根，所以可设特解为： y ∗ = ( a x + b ) cos ⁡ 2 x + ( c x + d ) sin ⁡ 2 x y^*=(a x+b) \cos 2 x+(c x+d) \sin 2 x y∗=(ax+b)cos2x+(cx+d)sin2x 代入方程得： ( − 3 a x − 3 b + 4 c ) cos ⁡ 2 x − ( 3 c x + 3 d + 4 a ) sin ⁡ 2 x = x cos ⁡ 2 x .  (-3 a x-3 b+4 c) \cos 2 x-(3 c x+3 d+4 a) \sin 2 x=x \cos 2 x \text {. } (−3ax−3b+4c)cos2x−(3cx+3d+4a)sin2x=xcos2x.  比较系数得： { − 3 a = 1 − 3 b + 4 c = 0 − 3 c = 0 − 3 d − 4 a = 0 \left\{\begin{array}{l} -3 a=1 \\ -3 b+4 c=0 \\ -3 c=0 \\ -3 d-4 a=0 \end{array}\right. ⎩ ⎨ ⎧​−3a=1−3b+4c=0−3c=0−3d−4a=0​ 由此解得： a = − 1 3 , b = 0 , c = 0 , d = 4 9 .  a=-\frac{1}{3}, b=0, c=0, d=\frac{4}{9} \text {. } a=−31​,b=0,c=0,d=94​.  于是特解为： y ∗ = − 1 3 x cos ⁡ 2 x + 4 9 sin ⁡ 2 x .  y^*=-\frac{1}{3} x \cos 2 x+\frac{4}{9} \sin 2 x \text {. } y∗=−31​xcos2x+94​sin2x.  其中代入方程这里要求一个二阶导，计算量非常大

首先解出两个齐次特解： r 2 − 1 = 0 ⟹ r 1 , 2 = ± 1 ⟹ y 1 = e x , y 2 = e − x r^2-1=0\Longrightarrow r_{1,2}=\pm 1\Longrightarrow y_1=\mathrm{e}^x,y_2=\mathrm{e}^{-x} \\ r2−1=0⟹r1,2​=±1⟹y1​=ex,y2​=e−x 再代公式解出非齐次特解： y ∗ = e x ∫ e x cos ⁡ 2 x e − x ⋅ ( e x e − x ) ′ d x + e − x ∫ e x cos ⁡ 2 x e x ⋅ ( e − x e x ) ′ d x = e x ∫ e x cos ⁡ 2 x e − x ⋅ 2 e 2 x   d x + e − x ∫ e x cos ⁡ 2 x − 2 e x e − 2 x   d x = 1 2 e x ∫ cos ⁡ 2 x   d x − 1 2 e − x ∫ e 2 x cos ⁡ 2 x   d x = 1 4 e x sin ⁡ 2 x − 1 4 e − x ∫ e 2 x   d ( sin ⁡ 2 x ) = 1 2 e − x ∫ sin ⁡ 2 x ⋅ e 2 x   d x = 1 8 e x ( sin ⁡ 2 x − cos ⁡ 2 x ) \begin{aligned} y^* & =\mathrm{e}^x \int \frac{\mathrm{e}^x \cos 2 x}{\mathrm{e}^{-x} \cdot\left(\frac{\mathrm{e}^x}{\mathrm{e}^{-x}}\right)^{\prime}} \mathrm{d} x+\mathrm{e}^{-x} \int \frac{\mathrm{e}^x \cos 2 x}{\mathrm{e}^x \cdot\left(\frac{\mathrm{e}^{-x}}{\mathrm{e}^x}\right)^{\prime}} \mathrm{d} x \\ & =\mathrm{e}^x \int \frac{\mathrm{e}^x \cos 2 x}{\mathrm{e}^{-x} \cdot 2 \mathrm{e}^{2 x}} \mathrm{~d} x+\mathrm{e}^{-x} \int \frac{\mathrm{e}^x \cos 2 x}{-2 \mathrm{e}^x \mathrm{e}^{-2 x}} \mathrm{~d} x \\ & =\frac{1}{2} \mathrm{e}^x \int \cos 2 x \mathrm{~d} x-\frac{1}{2} \mathrm{e}^{-x} \int \mathrm{e}^{2 x} \cos 2 x \mathrm{~d} x \\ & =\frac{1}{4} \mathrm{e}^x \sin 2 x-\frac{1}{4} \mathrm{e}^{-x} \int \mathrm{e}^{2 x} \mathrm{~d}(\sin 2 x) \\ & =\frac{1}{2} \mathrm{e}^{-x} \int \sin 2 x \cdot \mathrm{e}^{2 x} \mathrm{~d} x \\ & =\frac{1}{8} \mathrm{e}^x(\sin 2 x-\cos 2 x) \end{aligned} y∗​=ex∫e−x⋅(e−xex​)′excos2x​dx+e−x∫ex⋅(exe−x​)′excos2x​dx=ex∫e−x⋅2e2xexcos2x​ dx+e−x∫−2exe−2xexcos2x​ dx=21​ex∫cos2x dx−21​e−x∫e2xcos2x dx=41​exsin2x−41​e−x∫e2x d(sin2x)=21​e−x∫sin2x⋅e2x dx=81​ex(sin2x−cos2x)​ 其中最后一个等号可以套公式，具体公式及其推导如下： I 1 = ∫ e a x sin ⁡ b x   d x , I 2 = ∫ e a x cos ⁡ b x   d x I 1 = − 1 b ∫ e a x   d cos ⁡ b x = − 1 b ( e a x cos ⁡ b x − a ∫ e a x cos ⁡ b x   d x ) = a b I 2 − 1 b e a x cos ⁡ b x I 2 = 1 b ∫ e a x   d sin ⁡ b x = 1 b ( e a x sin ⁡ b x − a ∫ e a x sin ⁡ b x   d x ) = 1 b e a x sin ⁡ b x − a b I 1 { b I 1 − a I 2 = − e a x cos ⁡ b x a I 2 + b I 1 = e a x sin ⁡ b x ⟹ { I 1 = e a x a 2 + b 2 ( a sin ⁡ b x − b cos ⁡ b x ) + C I 2 = e a x a 2 + b 2 ( b sin ⁡ b x + a cos ⁡ b x ) + C \begin{aligned} & I_1=\int \mathrm{e}^{a x}\sin b x \mathrm{~d} x, I_2=\int \mathrm{e}^{a x}\cos b x \mathrm{~d} x \\ & I_1=-\frac{1}{b} \int \mathrm{e}^{a x} \mathrm{~d} \cos b x=-\frac{1}{b}\left(\mathrm{e}^{a x} \cos b x-a \int \mathrm{e}^{a x} \cos b x \mathrm{~d} x\right)=\frac{a}{b} I_2-\frac{1}{b} \mathrm{e}^{a x} \cos b x \\ & I_2=\frac{1}{b} \int \mathrm{e}^{a x} \mathrm{~d} \sin b x=\frac{1}{b}\left(\mathrm{e}^{a x} \sin b x-a \int \mathrm{e}^{a x} \sin b x \mathrm{~d} x\right)=\frac{1}{b} \mathrm{e}^{a x} \sin b x-\frac{a}{b} I_1 \\ & \left\{\begin{array} { l } { b I _ { 1 } - a I _ { 2 } = - \mathrm { e } ^ { a x } \operatorname { cos } b x } \\ { a I _ { 2 } + b I _ { 1 } = \mathrm { e } ^ { a x } \operatorname { sin } b x } \end{array} \Longrightarrow \left\{\begin{array}{l} I_1=\frac{\mathrm{e}^{a x}}{a^2+b^2}(a \sin b x-b \cos b x)+C \\ I_2=\frac{\mathrm{e}^{a x}}{a^2+b^2}(b \sin b x+a \cos b x)+C \end{array}\right.\right. \end{aligned} ​I1​=∫eaxsinbx dx,I2​=∫eaxcosbx dxI1​=−b1​∫eax dcosbx=−b1​(eaxcosbx−a∫eaxcosbx dx)=ba​I2​−b1​eaxcosbxI2​=b1​∫eax dsinbx=b1​(eaxsinbx−a∫eaxsinbx dx)=b1​eaxsinbx−ba​I1​{bI1​−aI2​=−eaxcosbxaI2​+bI1​=eaxsinbx​⟹{I1​=a2+b2eax​(asinbx−bcosbx)+CI2​=a2+b2eax​(bsinbx+acosbx)+C​​ 看起来计算量很大，但实际上仍然要比待定系数法计算量小，我们可以对比一下待定系数法的计算量。注意到 1 + 2 i 1+2 i 1+2i不是特征方程的根，所以可设特解为： y ∗ = e x ( a cos ⁡ 2 x + b sin ⁡ 2 x ) y^*=\mathrm{e}^x(a \cos 2 x+b \sin 2 x) y∗=ex(acos2x+bsin2x) 求一阶导和二阶导： y ∗ ′ = e x [ ( a + 2 b ) cos ⁡ 2 x + ( − 2 a + b ) sin ⁡ 2 x ] , y ∗ ′ ′ = e x [ ( − 3 a + 4 b ) cos ⁡ 2 x + ( − 4 a − 3 b ) sin ⁡ 2 x ] . \begin{aligned} & y^{* \prime}=\mathrm{e}^x[(a+2 b) \cos 2 x+(-2 a+b) \sin 2 x], \\ & y^{* \prime \prime}=\mathrm{e}^x[(-3 a+4 b) \cos 2 x+(-4 a-3 b) \sin 2 x] . \end{aligned} ​y∗′=ex[(a+2b)cos2x+(−2a+b)sin2x],y∗′′=ex[(−3a+4b)cos2x+(−4a−3b)sin2x].​ 代入方程得： 4 e x [ ( − a + b ) cos ⁡ 2 x − ( a + b ) sin ⁡ 2 x ] = e x cos ⁡ 2 x , 4 \mathrm{e}^x[(-a+b) \cos 2 x-(a+b) \sin 2 x]=\mathrm{e}^x \cos 2 x, 4ex[(−a+b)cos2x−(a+b)sin2x]=excos2x, 比较系数得： { − a + b = 1 4 , a + b = 0 , ⟹ { a = − 1 8 , b = 1 8 . \left\{\begin{array} { l } { - a + b = \frac { 1 } { 4 } , } \\ { a + b = 0 , } \end{array} \quad \Longrightarrow \left\{\begin{array}{l} a=-\frac{1}{8}, \\ b=\frac{1}{8} . \end{array}\right.\right. {−a+b=41​,a+b=0,​⟹{a=−81​,b=81​.​

于是特解为： y ∗ = 1 8 e x ( sin ⁡ 2 x − cos ⁡ 2 x ) y^*=\frac{1}{8} \mathrm{e}^x(\sin 2 x-\cos 2 x) y∗=81​ex(sin2x−cos2x) 其中代入方程这里的计算量非常大

[1]数学群管理员魏念辉给出此公式(与文献3本质上相同，但形式上更简单) [2]魏佬给出了云哥的证明过程(与文献3基本一致) [3]朱德刚. (2010). 二阶常系数非齐次线性微分方程的特解公式. 高等数学研究, 13(3), 2.