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2.3
2.求下列方程的解
(11)
x ( 4 y d x + 2 x d y ) + y 3 ( 3 y d x + 5 x d y ) = 0 x(4y\mathrm{d}x+2x\mathrm{d}y)+y^3(3y\mathrm{d}x+5x\mathrm{d}y)=0 x(4ydx+2xdy)+y3(3ydx+5xdy)=0
由于
M = 4 x y + 3 y 4 , N = 2 x 2 + 5 x y 3 M=4xy+3y^4,N=2x^2+5xy^3 M=4xy+3y4,N=2x2+5xy3
∂ M ∂ y = 4 x + 12 y 3 ≠ ∂ N ∂ x = 4 x + 5 y 3 \frac{\partial M}{\partial y}=4x+12y^3\neq \frac{\partial N}{\partial x}=4x+5y^3 ∂y∂M=4x+12y3=∂x∂N=4x+5y3
此方程不是恰当微分方程
因为 M , N M,N M,N 为多项式,设积分因子的形式为 μ = x m y n \mu=x^my^n μ=xmyn ,原方程乘上积分因子变为
( 4 x m + 1 y n + 1 + 3 x m y n + 4 ) d x + ( 2 x m + 2 y n + 5 x m + 1 y n + 3 ) d y = 0 (4x^{m+1}y^{n+1}+3x^my^{n+4})\mathrm{d}x+ (2x^{m+2}y^n+5x^{m+1}y^{n+3})\mathrm{d}y=0 (4xm+1yn+1+3xmyn+4)dx+(2xm+2yn+5xm+1yn+3)dy=0
它是恰当微分方程的充要条件为
4 ( n + 1 ) x m + 1 y n + 3 ( n + 4 ) x m y n + 3 = 2 ( m + 2 ) x m + 1 y n + 5 ( m + 1 ) x m y n + 3 4(n+1)x^{m+1}y^n+3(n+4)x^my^{n+3}=2(m+2)x^{m+1}y^n+5(m+1)x^my^{n+3} 4(n+1)xm+1yn+3(n+4)xmyn+3=2(m+2)xm+1yn+5(m+1)xmyn+3
对比系数项得
{ 4 ( n + 1 ) = 2 ( m + 2 ) 3 ( n + 4 ) = 5 ( m + 1 ) \begin{cases} 4(n+1)=2(m+2)\\ 3(n+4)=5(m+1) \end{cases} {4(n+1)=2(m+2)3(n+4)=5(m+1)
解得 m = 2 , n = 1 m=2,n=1 m=2,n=1
所以积分因子为 μ = x 2 y \mu=x^2y μ=x2y ,之后套公式就可以了,通解为
x 4 y 2 + x 3 y 5 = c x^4y^2+x^3y^5=c x4y2+x3y5=c
3.试导出方程 M ( x , y ) d x + N ( x , y ) d y = 0 M(x,y)\mathrm{d}x+N(x,y)\mathrm{d}y=0 M(x,y)dx+N(x,y)dy=0 分别具有形为 μ ( x + y ) \mu(x+y) μ(x+y) 和 μ ( x y ) \mu(xy) μ(xy) 的积分因子的充要条件
假设方程具有 μ ( x + y ) \mu(x+y) μ(x+y) 的积分因子
令 z = x + y z=x+y z=x+y ,则 μ ( x + y ) = μ ( z ) \mu(x+y)=\mu(z) μ(x+y)=μ(z) ,且
∂ μ ∂ x = ∂ μ ∂ y = ∂ μ ∂ z \frac{\partial\mu}{\partial x}= \frac{\partial\mu}{\partial y}= \frac{\partial\mu}{\partial z} ∂x∂μ=∂y∂μ=∂z∂μ
所以由
∂ ( μ M ) ∂ y = ∂ ( μ N ) ∂ x \frac{\partial(\mu M)}{\partial y}=\frac{\partial(\mu N)}{\partial x} ∂y∂(μM)=∂x∂(μN)
N ∂ μ ∂ x − M ∂ μ ∂ y = ( ∂ M ∂ y − ∂ N ∂ x ) μ N\frac{\partial\mu}{\partial x}-M\frac{\partial\mu}{\partial y}= \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)\mu N∂x∂μ−M∂y∂μ=(∂y∂M−∂x∂N)μ
( N − M ) d μ d z = ( ∂ M ∂ y − ∂ N ∂ x ) μ ( z ) (N-M)\frac{\mathrm{d}\mu}{\mathrm{d}z}= \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right) \mu(z) (N−M)dzdμ=(∂y∂M−∂x∂N)μ(z)
即
d μ μ ( z ) = ∂ M ∂ y − ∂ N ∂ x N − M d z \frac{\mathrm{d}\mu}{\mu(z)}= \frac{\frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}}{N-M}\mathrm{d}z μ(z)dμ=N−M∂y∂M−∂x∂Ndz
当
∂ M ∂ y − ∂ N ∂ x N − M = f ( z ) = f ( x + y ) \frac{\frac{\partial M}{\partial y} -\frac{\partial N}{\partial x}}{N-M}=f(z)=f(x+y) N−M∂y∂M−∂x∂N=f(z)=f(x+y)
成立时,可以解出 μ \mu μ ,这就是方程具有 μ ( x + y ) \mu(x+y) μ(x+y) 的积分因子的充要条件
没有分充分性必要性一一证明,凑合用吧
同理,令 w = x y w=xy w=xy ,则
∂ μ ∂ x = y ∂ μ ∂ w , ∂ μ ∂ y = x ∂ μ ∂ w , \frac{\partial\mu}{\partial x}=y\frac{\partial\mu}{\partial w}, \frac{\partial\mu}{\partial y}=x\frac{\partial\mu}{\partial w}, ∂x∂μ=y∂w∂μ,∂y∂μ=x∂w∂μ,
由判定恰当微分方程的式子
( y N − x M ) d μ d w = ( ∂ M ∂ y − ∂ N ∂ x ) μ ( w ) (yN-xM)\frac{\mathrm{d}\mu}{\mathrm{d}w}= \left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)\mu(w) (yN−xM)dwdμ=(∂y∂M−∂x∂N)μ(w)
方程具有 μ ( x y ) \mu(xy) μ(xy) 的积分因子的充要条件
∂ M ∂ y − ∂ N ∂ x y N − x M = g ( x y ) \frac{\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}}{yN-xM} =g(xy) yN−xM∂y∂M−∂x∂N=g(xy)