Maple/常微分方程

線性常微分方程

dsolve(diff(y(x), x, x) = y(x))

$y(x)=C1\ast exp(x)+C2\ast exp(-x)$

f := diff(y(x), x, x)+a^2*y(x);

$y(x){=}_{C}1\ast sin(a\ast x){+}_{C}2\ast cos(a\ast x)$

$a\ast \frac{d^{2}x}{d{x}^2}+b\ast \frac{dy}{dx}+cy+d=0$

Maple:

$f:=a\ast (diff(y(x),x,x))+b\ast (diff(y(x),x))+cy+d=0$

dsolve(f);

$y(x)=\frac{-a\ast exp(-b\ast x/a){\ast }_{C}1}{b}-\frac{(cy+d)\ast x}{b}{+}_{C}2$

dsolve(diff(y(x), x, x, x, x) = y(x));

y(x) = _C1*exp(x)+_C2*exp(-x)+_C3*sin(x)+_C4*cos(x)

dsolve(diff(y(x), x, x, x, x, x, x, x) = diff(y(x), x, x))

y(x) = _C1+_C2*x+_C3*exp(x)-_C4*exp((-1/4-(1/4)*sqrt(5))*x)*sin((1/4)*sqrt(2)*sqrt(5-sqrt(5))*x)-_C5*exp((-1/4+(1/4)*sqrt(5))*x)*sin((1/4)*sqrt(2)*sqrt(5+sqrt(5))*x)+_C6*exp((-1/4-(1/4)*sqrt(5))*x)*cos((1/4)*sqrt(2)*sqrt(5-sqrt(5))*x)+_C7*exp((-1/4+(1/4)*sqrt(5))*x)*cos((1/4)*sqrt(2)*sqrt(5+sqrt(5))*x)

$f:=a\ast \frac{d^{3}y}{d{x}^3}+b\ast \frac{d^{2}y}{d{x}^2}+c\ast \frac{dy}{dx}+d\ast y=e$

Maple 式：

$f:=a\ast (diff(y(x),x,x,x))+b\ast (diff(y(x),x,x))+c\ast (diff(y(x),x))+d\ast y(x)=e$

dsolve(f);

y(x) = e/d+_C1*exp((1/6)*((36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(2/3)-12*c*a+4*b^2-2*b*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3))*x/(a*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3)))+_C2*exp(-(1/12*I)*(-I*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(2/3)+(12*I)*c*a-(4*I)*b^2-(4*I)*b*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3)+sqrt(3)*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(2/3)+12*sqrt(3)*c*a-4*sqrt(3)*b^2)*x/(a*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3)))+_C3*exp((1/12*I)*(I*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(2/3)-(12*I)*c*a+(4*I)*b^2+(4*I)*b*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3)+sqrt(3)*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(2/3)+12*sqrt(3)*c*a-4*sqrt(3)*b^2)*x/(a*(36*c*b*a-108*d*a^2-8*b^3+12*sqrt(3)*sqrt(4*c^3*a-c^2*b^2-18*c*b*a*d+27*d^2*a^2+4*d*b^3)*a)^(1/3)))

分離變數型微分方程

$\frac{d(x)}{dx}=f(t)\ast g(x)$

• Maple 可以直接得出解答，無需用戶作分離變數步驟。
• $\frac{dy(x)}{dx}=\frac{1+y(x{)}^2}{1+x^2}$
• f := diff(y(x), x) = (1+y(x)^2)/(1+x^2)；
• dsolve(f);
• $y(x)=tan(arctan(x){+}_{C}1)$

$\frac{dy(x)}{dx}=\frac{(1+y(x{)}^2)\ast (1+1/x)}{1+x^2}$

• f := diff(y(x), x) = (1+y(x)^2)*(1+1/x)/(1+x^2)
• dsolve(f);

• $y(x)=tan(ln(x)-(1/2)\ast ln(1+x^2)+arctan(x){+}_{C}1)$

$f:=\frac{d(x(t)}{dt}=\frac{(x(t{)}^2+t^2)}{t\ast x(t)}$

歐拉型微分方程

$\frac{dx}{dt}=\frac{P(x,t)}{Q(x,t)}$

• f := diff(x(t), t) = (x(t)^2+t^2)/(t*x(t))

• dsolve(f);

• $x(t)=\sqrt{2\ast ln(t){+}_{C}1}\ast t$,
• $x(t)=-\sqrt{2\ast ln(t){+}_{C}1}\ast t$

$f:=\frac{dx}{dt}=\frac{x+t}{5x+t}$

• dsolve(f);

• $x(t)=-(1/5)\ast t-(1/5)\ast \sqrt{t^2-10\ast t+10{\ast }_{C}1}$

• $x(t)=-(1/5)\ast t+(1/5)\ast \sqrt{t^2-10\ast t+10{\ast }_{C}1}$

伯努利微分方程

$\frac{dx}{dt}$$=A(t)\ast x+B(t)\ast x^a$

$\frac{dx(t)}{dt}=5\ast x(t)/t+t\ast \sqrt{x(t)}$

f := diff(x(t), t) = 5*x(t)/t+t*x(t)^(1/2)

• dsolve(f);

• $\sqrt{x(t)}+t^2-t^{(}5/2){\ast }_{C}1=0$

Riccati 型微分方程

$A(t)x^3+B(t)x+C$

$f:=\frac{d(x(t)}{dt}=sin(t)\ast x(t{)}^2+2\ast \frac{sin(t)}{cos(t{)}^2}$

• f := diff(x(t), t) = sin(t)*x(t)^2+2*sin(t)/cos(t)^2

• dsolve(f);

• $x(t)=-(1/14)\ast (-\sqrt{(}7)+$$7\ast tan((1/2\ast (ln(cos(t)){+}_{C}1))\ast \sqrt{(}7)))\ast \sqrt{(}7)/cos(t)$

$f:=\frac{d(x(t)}{dt}=t^5\ast x(t{)}^2+5\ast x(t)/t+6/t^2$

• f := diff(x(t), t) = t^5*x(t)^2+5*x(t)/t+6/t^2

• dsolve(f);

• $x(t)=\frac{-\sqrt{(}6)\ast (BesselJ(6/5,(2/5)\ast \sqrt{(}6)\ast t^{(}5/2)){+}_{C}1\ast BesselY(6/5,(2/5)\ast \sqrt{(}6)\ast t^{(}5/2)))}{(t^{(}7/2)\ast {(}_{C}1\ast BesselY(11/5,(2/5)\ast \sqrt{(}6)\ast t^{(}5/2))+BesselJ(11/5,(2/5)\ast \sqrt{(}6)\ast t^{(}5/2))))}$

非線性常微分方程

$f=(1-x^2)\ast \frac{d^{2}y(x)}{d{x}^2}-2\ast x\ast \frac{dy(x)}{dx}+l(l+1)$

f := (1-x^2)*(diff(y(x), x, x))-2*x*(diff(y(x), x))+l*(l+1);

> dsolve(f);

y(x) = (1/2)*ln(x-1)*_C1+(1/2)*ln(x-1)*l^2+(1/2)*ln(x-1)*l-(1/2)*ln(x+1)*_C1+(1/2)*ln(x+1)*l^2+(1/2)*ln(x+1)*l+_C2

$f=x^2\ast \frac{d^{2}y(x)}{d{x}^2}+x\ast \frac{dy(x)}{dx}+(x^2-v^2)\ast y$

f := x^2*(diff(y(x), x, x))+x*(diff(y(x), x))+(x^2-v^2)*y(x)

dsolve(f);

y(x) = _C1*BesselJ(v, x)+_C2*BesselY(v, x)

$f=(1-x^2)\ast \frac{d^{2}y(x)}{d{x}^2}-2\ast x\ast \frac{dy(x)}{dx}+l(l+1)\ast y$

f := (1-x^2)*(diff(y(x), x, x))-2*x*(diff(y(x), x))+l(l+1)*y(x);

dsolve(f);

y(x) = _C1*LegendreP((1/2)*sqrt(1+4*l(l+1))-1/2, x)+_C2*LegendreQ((1/2)*sqrt(1+4*l(l+1))-1/2, x)

$\frac{dy(x)}{d^{2}x}-x\ast \frac{dy(x)}{dx}+n\ast y(x)$

f := diff(y(x), x, x)-x*(diff(y(x), x))+n*y(x);

dsolve(f);

y(x) = _C1*KummerM(1/2-(1/2)*n, 3/2, (1/2)*x^2)*x+_C2*KummerU(1/2-(1/2)*n, 3/2, (1/2)*x^2)*x

$x\ast \frac{d^{3}y(x)}{d^{3}x}+x\ast \frac{dy(x)}{dx}+y(x)-1$

f := x*(diff(y(x), x, x, x))+x*(diff(y(x), x))+y(x)-1

dsolve(f);

y(x) = x*BesselJ(1, x)*_C3+x*BesselY(1, x)*_C2+(1/4)*π*x*(-BesselJ(1, x)*BesselY(0, x)+BesselY(1, x)*BesselJ(0, x))*(Pi*x*_C1*StruveH(-1, x)-2)

$\frac{d^{2}y(x)}{d^{2}x}+x^3\ast \frac{dy(x)}{dx}+y(x)-1$

y(x) = exp(-(1/4)*x^4)*HeunB(-1/2, 0, -3/2, 1, -(1/2)*x^2)*_C2+exp(-(1/4)*x^4)*HeunB(1/2, 0, -3/2, 1, -(1/2)*x^2)*x*_C1+1

$\frac{d^{2}y(x)}{d^{2}x}+(1-x^2)\ast \frac{dy(x)}{dx}+x\ast y-1$

f := diff(y(x), x, x)+(1-x^2)*(diff(y(x), x))+x*y(x)-1

dsolve(f);

y(x) = HeunT(0, 6, -3^(1/3), (1/3)*3^(2/3)*x)*_C2+HeunT(0, -6, -3^(1/3), -(1/3)*3^(2/3)*x)*exp((1/3)*x*(-3+x^2))*_C1+x;

$\frac{d^2(x)}{d^{3}x}-cos(x)\ast y(x)-1$

f := diff(y(x), x, x)-cos(x)*y(x)-1

y(x) = MathieuC(0, 2, (1/2)x| _C2 + MathieuS(0, 2, (1/2)x) _C1 - 2 | |

 /        /      1  \\//        /      1  \              /      
 |MathieuS|0, 2, - x|| |MathieuC|0, 2, - x| MathieuSPrime|0, 2, 
 \        \      2  // \        \      2  /              \      

                                                        \ 
 1  \           /      1  \              /      1  \\   | 
 - x| - MathieuS|0, 2, - x| MathieuCPrime|0, 2, - x|| dx| 
 2  /           \      2  /              \      2  //   | 
                                                        / 

                         /  /                                    
         /      1  \     | |  /        /      1  \\//        /   
 MathieuC|0, 2, - x| + 2 | |  |MathieuC|0, 2, - x|| |MathieuC|0, 
         \      2  /     | |  \        \      2  // \        \   
                         \/                                      

    1  \              /      1  \
 2, - x| MathieuSPrime|0, 2, - x|
    2  /              \      2  /

                                                    \          
            /      1  \              /      1  \\   |         /
  - MathieuS|0, 2, - x| MathieuCPrime|0, 2, - x|| dx| MathieuS|
            \      2  /              \      2  //   |         \
                                                    /          

       1  \
 0, 2, - x|
       2  /

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