1)
y ′ + 3 y = s i n ( x ) {\displaystyle y'+3y=sin(x)\,\!}
步骤 1:找到 e ∫ P ( x ) d x {\displaystyle e^{\int P(x)dx}}
∫ 3 d x = 3 x + C {\displaystyle \int 3dx=3x+C}
e ∫ P ( x ) d x = C e 3 x {\displaystyle e^{\int P(x)dx}=Ce^{3x}}
令 C=1,得到 e 3 x {\displaystyle e^{3x}}
步骤 2:乘以整个式子
e 3 x y ′ + e 3 x 3 y = e 3 x s i n ( x ) {\displaystyle e^{3x}y'+e^{3x}3y=e^{3x}sin(x)\,\!}
步骤 3:识别出左边是 d d x e ∫ P ( x ) d x y {\displaystyle {\frac {d}{dx}}e^{\int P(x)dx}y}
d d x e 3 x y = e 3 x s i n ( x ) {\displaystyle {\frac {d}{dx}}e^{3x}y=e^{3x}sin(x)}
步骤 4:积分
∫ ( d d x e 3 x y ) d x = ∫ e 3 x s i n ( x ) d x {\displaystyle \int ({\frac {d}{dx}}e^{3x}y)dx=\int e^{3x}sin(x)dx}
e 3 x y = e 3 x ( 3 s i n ( x ) − c o s ( x ) ) 10 + C {\displaystyle e^{3x}y={\frac {e^{3x}(3sin(x)-cos(x))}{10}}+C}
步骤 5:解出 y
y = 3 s i n ( x ) − c o s ( x ) 10 + C e 3 x {\displaystyle y={\frac {3sin(x)-cos(x)}{10}}+{\frac {C}{e^{3x}}}}
2)
y ′ + 1 x + 3 y = 7 x 2 + 4 x {\displaystyle y'+{\frac {1}{x+3}}y=7x^{2}+4x}
∫ d x x + 3 = l n ( x + 3 ) + C {\displaystyle \int {\frac {dx}{x+3}}=ln(x+3)+C}
e ∫ P ( x ) d x = C x + 3 C {\displaystyle e^{\int P(x)dx}=Cx+3C}
令 C=1,得到 x + 3 {\displaystyle x+3}
( x + 3 ) y ′ + ( x + 3 ) y = ( x + 3 ) ( 7 x 2 + 4 x ) {\displaystyle (x+3)y'+(x+3)y=(x+3)(7x^{2}+4x)\,\!}
d d x ( x + 3 ) y = ( x + 3 ) ( 7 x 2 + 4 x ) {\displaystyle {\frac {d}{dx}}(x+3)y=(x+3)(7x^{2}+4x)}
∫ ( d d x ( x + 3 ) y ) d x = ∫ ( x + 3 ) ( 7 x 2 + 4 x ) d x {\displaystyle \int ({\frac {d}{dx}}(x+3)y)dx=\int (x+3)(7x^{2}+4x)dx}
( x + 3 ) y = 7 x 4 4 + 25 x 3 3 + 6 x 2 + C {\displaystyle (x+3)y={\frac {7x^{4}}{4}}+{\frac {25x^{3}}{3}}+6x^{2}+C}
y = 7 x 4 4 + 25 x 3 3 + 6 x 2 + C x + 3 {\displaystyle y={\frac {{\frac {7x^{4}}{4}}+{\frac {25x^{3}}{3}}+6x^{2}+C}{x+3}}}
3)
( x 4 e x − 2 m x y 2 ) d x + 2 m x 2 y d y {\displaystyle (x^{4}e^{x}-2mxy^{2})dx+2mx^{2}ydy}
步骤 1:重新排列
2 y d y d x − 2 y 2 x = − x 2 e x {\displaystyle 2y{\dfrac {dy}{dx}}-{\frac {2y^{2}}{x}}=-x^{2}e^{x}}
步骤 2:替换 z = y 2 ⟹ d z d x = 2 y d y d x {\displaystyle z=y^{2}\implies {\dfrac {dz}{dx}}=2y{\dfrac {dy}{dx}}}
d z d x − 2 z x = − x 2 e x {\displaystyle {\dfrac {dz}{dx}}-2{\frac {z}{x}}=-x^{2}e^{x}}
步骤 3:找出 e ∫ P ( x ) d x {\displaystyle e^{\int P(x)dx}}
积分因子 = 1 x 2 {\displaystyle ={\frac {1}{x^{2}}}}
步骤 4:求解 y
y ( x ) = x 2 ∫ − x 2 e x d x m x 2 = x 2 ∫ − e x m d x = − x 2 e x m + C x 2 {\displaystyle y(x)=x^{2}\int {\frac {-x^{2}e^{x}dx}{mx^{2}}}=x^{2}\int {\frac {-e^{x}}{m}}dx={\frac {-x^{2}e^{x}}{m}}+Cx^{2}}