【1989-12-3 分】 设线性无关的函数 y 1 , y 2 , y 3 \displaystyle y_{1}, y_{2}, y_{3} y 1 , y 2 , y 3 y ′ ′ + p ( x ) y ′ + q ( x ) y = f ( x ) \displaystyle y^{\prime \prime}+p(x) y'+q(x) y=f(x) y ′′ + p ( x ) y ′ + q ( x ) y = f ( x ) c 1 , c 2 \displaystyle c_{1}, c_{2} c 1 , c 2 c 1 y 1 + c 2 y 2 + y 3 \displaystyle c_{1} y_{1}+c_{2} y_{2}+y_{3} c 1 y 1 + c 2 y 2 + y 3 c 1 y 1 + c 2 y 2 − ( c 1 + c 2 ) y 3 \displaystyle c_{1} y_{1}+c_{2} y_{2}-\left(c_{1}+c_{2}\right) y_{3} c 1 y 1 + c 2 y 2 − ( c 1 + c 2 ) y 3 c 1 y 1 + c 2 y 2 − ( 1 − c 1 − c 2 ) y 3 \displaystyle c_{1} y_{1}+c_{2} y_{2}-\left(1-c_{1}-c_{2}\right) y_{3} c 1 y 1 + c 2 y 2 − ( 1 − c 1 − c 2 ) y 3 c 1 y 1 + c 2 y 2 + ( 1 − c 1 − c 2 ) y 3 \displaystyle c_{1} y_{1}+c_{2} y_{2}+\left(1-c_{1}-c_{2}\right) y_{3} c 1 y 1 + c 2 y 2 + ( 1 − c 1 − c 2 ) y 3
【1997-2-5 分】 已知 y 1 = x e x + e 2 x \displaystyle y_{1}=x e^{x}+e^{2 x} y 1 = x e x + e 2 x y 2 = x e x + e − x \displaystyle y_{2}=x e^{x}+e^{-x} y 2 = x e x + e − x y 3 = x e x + e 2 x − e − x \displaystyle y_{3}=x e^{x}+e^{2 x}-e^{-x} y 3 = x e x + e 2 x − e − x
【2006-3-4 分】 设非齐次线性微分方程 y ′ + P ( x ) y = Q ( x ) \displaystyle y'+P(x) y=Q(x) y ′ + P ( x ) y = Q ( x ) y 1 ( x ) \displaystyle y_{1}(x) y 1 ( x ) y 2 ( x ) \displaystyle y_{2}(x) y 2 ( x ) C [ y 1 ( x ) − y 2 ( x ) ] \displaystyle C\left[y_{1}(x)-y_{2}(x)\right] C [ y 1 ( x ) − y 2 ( x ) ] y 1 ( x ) + C [ y 1 ( x ) − y 2 ( x ) ] \displaystyle y_{1}(x)+C\left[y_{1}(x)-y_{2}(x)\right] y 1 ( x ) + C [ y 1 ( x ) − y 2 ( x ) ] C [ y 1 ( x ) + y 2 ( x ) ] \displaystyle C\left[y_{1}(x)+y_{2}(x)\right] C [ y 1 ( x ) + y 2 ( x ) ] y 1 ( x ) + C [ y 1 ( x ) + y 2 ( x ) ] \displaystyle y_{1}(x)+C\left[y_{1}(x)+y_{2}(x)\right] y 1 ( x ) + C [ y 1 ( x ) + y 2 ( x ) ]
【2010-23-4 分】 设 y 1 , y 2 \displaystyle y_{1}, y_{2} y 1 , y 2 y ′ + p ( x ) y = q ( x ) \displaystyle y'+p(x) y=q(x) y ′ + p ( x ) y = q ( x ) λ y 1 + μ y 2 \displaystyle \lambda y_{1}+\mu y_{2} λ y 1 + μ y 2 λ y 1 − μ y 2 \displaystyle \lambda y_{1}-\mu y_{2} λ y 1 − μ y 2 λ = 1 2 , μ = 1 2 \displaystyle \lambda=\dfrac{1}{2}, \mu=\dfrac{1}{2} λ = 2 1 , μ = 2 1 λ = − 1 2 , μ = − 1 2 \displaystyle \lambda=-\dfrac{1}{2}, \mu=-\dfrac{1}{2} λ = − 2 1 , μ = − 2 1 λ = 2 3 , μ = 1 3 \displaystyle \lambda=\dfrac{2}{3}, \mu=\dfrac{1}{3} λ = 3 2 , μ = 3 1 λ = 2 3 , μ = 2 3 \displaystyle \lambda=\dfrac{2}{3}, \mu=\dfrac{2}{3} λ = 3 2 , μ = 3 2
【2013-1-4 分】 已知 y 1 = e 3 x − x e 2 x , y 2 = e x − x e 2 x , y 3 = − x e 2 x \displaystyle y_{1}=e^{3 x}-x e^{2 x},y_{2}=e^{x}-x e^{2 x},y_{3}=-x e^{2 x} y 1 = e 3 x − x e 2 x , y 2 = e x − x e 2 x , y 3 = − x e 2 x y = \displaystyle y= y =
【2013-2-4 分】 已知 y 1 = e 3 x − x e 2 x , y 2 = e x − x e 2 x , y 3 = − x e 2 x \displaystyle y_{1}=e^{3 x}-x e^{2 x},y_{2}=e^{x}-x e^{2 x},y_{3}=-x e^{2 x} y 1 = e 3 x − x e 2 x , y 2 = e x − x e 2 x , y 3 = − x e 2 x y ∣ x = 0 = 0 \displaystyle y|_{x=0}=0 y ∣ x = 0 = 0 y ′ ∣ x = 0 = 1 \displaystyle y'|_{x=0}=1 y ′ ∣ x = 0 = 1 y = \displaystyle y= y =
【2016-2-4 分】 y = ( 1 + x 2 ) 2 + 1 + x 2 \displaystyle y=(1+x^{2})^{2}+\sqrt{1+x^{2}} y = ( 1 + x 2 ) 2 + 1 + x 2 y = ( 1 + x 2 ) 2 − 1 + x 2 \displaystyle y=(1+x^{2})^{2}-\sqrt{1+x^{2}} y = ( 1 + x 2 ) 2 − 1 + x 2 y ′ + p ( x ) y = q ( x ) \displaystyle y'+p(x) y=q(x) y ′ + p ( x ) y = q ( x ) q ( x ) = \displaystyle q(x)= q ( x ) = 3 x ( 1 + x 2 ) \displaystyle 3 x\left(1+x^{2}\right) 3 x ( 1 + x 2 ) − 3 x ( 1 + x 2 ) \displaystyle -3 x\left(1+x^{2}\right) − 3 x ( 1 + x 2 ) x 1 + x 2 \displaystyle \dfrac{x}{1+x^{2}} 1 + x 2 x − x 1 + x 2 \displaystyle -\dfrac{x}{1+x^{2}} − 1 + x 2 x
【2016-2-4 分】 以 y = x 2 − e x \displaystyle y=x^{2}-e^{x} y = x 2 − e x y = x 2 \displaystyle y=x^{2} y = x 2
【2019-23-4 分】 已知微分方程 y ′ ′ + a y ′ + b y = c e x \displaystyle y^{\prime \prime}+a y'+b y=c e^{x} y ′′ + a y ′ + b y = c e x y = ( C 1 + C 2 x ) e − x + e x \displaystyle y=(C_{1}+C_{2} x) e^{-x}+e^{x} y = ( C 1 + C 2 x ) e − x + e x
【2021-2-5 分】 微分方程 y ′ ′ ′ − y = 0 \displaystyle y^{\prime \prime \prime}-y=0 y ′′′ − y = 0 y = \displaystyle y= y =