一阶微分方程

常数变易法

考虑以下一阶线性微分方程

$${\displaystyle \frac{\text{d}y}{\text{d}x}}=P(x)y+Q(x)$$

先解齐次方程

$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& =P(x)y\\ {\displaystyle \frac{\text{d}y}{y}}& =P(x)\text{d}x\\ \ln y& =P_1(x)+C\\ y& =C\exp (P_1(x))\end{array}$$

其中

$$P_1(x)=\int P(x)\text{d}x$$

根据常数变易法，对于原非齐次方程有

$$y=C(x)\exp (P_1(x))$$

即将 $C$ 改为关于 $x$ 的函数 $C(x)$。

将 $y$ 代回原方程即可解出 $C(x)$。

这个方法可以拓展到非线性微分方程。

P6613

考虑以下方程

$$\frac{\text{d}F(x)}{\text{d}x}\equiv A(x){\text{e}}^{F(x)-1}+B(x)(\mathrm{mod}{x}^n)$$

设 $y=F(x)$，先解齐次方程

$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& =A(x)\exp (y-1)\\ {\displaystyle \frac{\text{d}y}{\exp (y-1)}}& =A(x)\text{d}x\\ \exp (1-y)\text{d}y& =A(x)\text{d}x\\ -\exp (1-y)& =A_1(x)-C\\ \exp (1-y)& =C-A_1(x)\\ 1-y& =\ln (C-A_1(x))\\ y& =1-\ln (C-A_1(x))\\ y& =1+\ln {\displaystyle \frac{1}{C-A_1(x)}}\end{array}$$

其中

$$A_1(x)=\int A(x)\text{d}x$$

根据常数变易法，对于原非齐次方程有

$$\begin{array}{rl}y& =1-\ln (C(x)-A_1(x))\\ y& =1+\ln {\displaystyle \frac{1}{C(x)-A_1(x)}}\end{array}$$

代回原方程

$$\begin{array}{rl}{\displaystyle \frac{\text{d}y}{\text{d}x}}& =A(x)\exp (y-1)+B(x)\\ -{\displaystyle \frac{C^{\prime }(x)-A(x)}{C(x)-A_1(x)}}& ={\displaystyle \frac{A(x)}{C(x)-A_1(x)}}+B(x)\\ {\displaystyle \frac{C^{\prime }(x)}{C(x)-A_1(x)}}& =-B(x)\\ C^{\prime }(x)& =-B(x)C(x)+A_1(x)B(x)\end{array}$$

设 $z=C(x)$

$$z^{\prime }=-B(x)z+A_1(x)B(x)$$

解此方程，求出通解

$$z=C_1(x)\exp (-B_1(x))$$

其中

$$B_1(x)=\int B(x)\text{d}x$$

代入得

$$\begin{array}{rl}{z}^{\prime }& =-B(x)z+A_1(x)B(x)\\ C_1^{\prime }(x)\exp (-B_1(x))-C_1(x)\exp (-B_1(x))B(x)& =-B(x)C_1(x)\exp (-B_1(x))+A_1(x)B(x)\\ C_1^{\prime }(x)\exp (-B_1(x))& =A_1(x)B(x)\\ C_1(x)& =\int A_1(x)B(x)\exp (B_1(x))\text{d}x+C\end{array}$$

继续回代

$$C(x)=(\int A_1(x)B(x)\exp (B_1(x))\text{d}x+C)\exp (-B_1(x))$$

$$\begin{array}{rl}F(x)& =1-\ln ((\int A_1(x)B(x)\exp (B_1(x))\text{d}x+C)\exp (-B_1(x))-A_1(x))\\ & =1+B_1(x)-\ln (\int A_1(x)B(x)\exp (B_1(x))\text{d}x-A_1(x)\exp (B_1(x))+C)\end{array}$$

当 $x=0$，$F(x)=1$，故取 $C=1$。

注意到

$$B(x)\exp (B_1(x))=\exp (B_1(x){)}^{\prime }$$

以及

$$A_1(x)\exp (B_1(x){)}^{\prime }=(A_1(x)\exp (B_1(x)){)}^{\prime }-A(x)\exp (B_1(x))$$

所以

$$F(x)=1+B_1(x)-\ln (1-\int A(x)\exp (B_1(x))\text{d}x)$$

直接计算即可。

Code

int main(){
	int n;
	scanf("%d",&n);
	n++;
	for(int i=0;i<=n-1;i++){
		scanf("%llu",&a[i]);
	}
	for(int i=0;i<=n-1;i++){
		scanf("%llu",&b[i]);
		b1[i]=b[i];
	}
	inv[1]=1;
	for(int i=2;i<=n;i++){
		inv[i]=1ll*(mod-mod/i)*inv[mod%i]%mod;
	}
	ITG(b1,n);
	EXP(b1,c,n);
	CPY(ans,b1,n);
	ans[0]=(ans[0]+1)%mod;
	CLR(b1,n);
	MUL(a,c,n<<1);
	CLR(a+n,n);
	ITG(a,n);
	a[0]=(1-a[0]+mod)%mod;
	for(int i=1;i<=n-1;i++){
		a[i]=(mod-a[i])%mod;
	}
	LN(a,c,n);
	DEC(ans,c,n);
	for(int i=0;i<=n-1;i++){
		printf("%llu ",ans[i]);
	}
}