查看“︁Maple/常微分方程”︁的源代码

==线性常微分方程==
dsolve(diff(y(x), x, x) = y(x))

<math>y(x) = C1*exp(x)+C2*exp(-x)</math>

f := diff(y(x), x, x)+a^2*y(x);

<math>y(x) = _C1*sin(a*x)+_C2*cos(a*x)</math>

<math>a*\frac{d^2x}{dx^2}+b*\frac{dy}{dx}+cy+d=0</math>

Maple:

<math>f := a*(diff(y(x), x, x))+b*(diff(y(x), x))+cy+d = 0</math>

dsolve(f);

<math>y(x) =\frac{ -a*exp(-b*x/a)*_C1}{b}-\frac{(cy+d)*x}{b}+_C2</math>

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)


<math>f:=a*\frac{d^3y}{dx^3}+b*\frac{d^2y}{dx^2}+c*\frac{dy}{dx}+d*y=e</math>

Maple 式：

<math>f := a*(diff(y(x), x, x, x))+b*(diff(y(x), x, x))+c*(diff(y(x), x))+d*y(x) = e</math>

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)))

==分离变数型微分方程==
<math> \frac{d(x)}{dx}=f(t)*g(x)</math>

* Maple 可以直接得出解答，无需用户作分离变数步骤。
* 
* <math>\frac{dy(x)}{dx}=\frac{1+y(x)^2}{1+x^2}</math>
* 
* f := diff(y(x), x) = (1+y(x)^2)/(1+x^2)；
* 
* dsolve(f);
* 
* <math>y(x) = tan(arctan(x)+_C1)</math>

<math>\frac{dy(x)}{dx}=\frac{(1+y(x)^2)*(1+1/x)}{1+x^2}</math>


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

*<math>y(x) = tan(ln(x)-(1/2)*ln(1+x^2)+arctan(x)+_C1)</math>

<math>f := \frac{d(x(t)}{dt} = \frac{(x(t)^2+t^2)}{t*x(t)}</math>
===欧拉型微分方程===

<math>\frac{dx}{dt}=\frac{P(x,t)}{Q(x,t)}</math>



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

*dsolve(f);

*<math>x(t) = \sqrt{2*ln(t)+_C1}*t</math>,
*<math> x(t) = -\sqrt{2*ln(t)+_C1}*t</math>

<math>f:=\frac{dx}{dt}=\frac{x+t}{5x+t}</math>

*dsolve(f);

*<math>x(t) = -(1/5)*t-(1/5)*\sqrt{t^2-10*t+10*_C1}</math>

*<math>x(t) = -(1/5)*t+(1/5)*\sqrt{t^2-10*t+10*_C1}</math>

==伯努利微分方程==
<math>\frac{dx}{dt}</math><math>=A(t)*x+B(t)*x^a</math>

<math>\frac{dx(t)}{dt}= 5*x(t)/t+t*\sqrt{x(t)}</math>

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

*dsolve(f);

*<math>\sqrt{x(t)}+t^2-t^(5/2)*_C1 = 0</math>

==Riccati 型微分方程==

<math>A(t)x^3+B(t)x+C</math>

<math>f := \frac{d(x(t)}{dt} = sin(t)*x(t)^2+2*\frac{sin(t)}{cos(t)^2}</math>

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

*dsolve(f);

*<math>x(t) = -(1/14)*(-\sqrt(7)+</math><math>7*tan((1/2*(ln(cos(t))+_C1))*\sqrt(7)))*\sqrt(7)/cos(t)</math>

<math>f := \frac{d(x(t)}{dt} = t^5*x(t)^2+5*x(t)/t+6/t^2</math>


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

*dsolve(f);

*<math>x(t) =\frac{ -\sqrt(6)*(BesselJ(6/5, (2/5)*\sqrt(6)*t^(5/2))+_C1*BesselY(6/5, (2/5)*\sqrt(6)*t^(5/2)))}{(t^(7/2)*(_C1*BesselY(11/5, (2/5)*\sqrt(6)*t^(5/2))+BesselJ(11/5, (2/5)*\sqrt(6)*t^(5/2))))}</math>

==非线性常微分方程==
<math> f=(1-x^2)*\frac{d^2y(x)}{dx^2}-2*x*\frac{dy(x)}{dx}+l(l+1)</math>

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


<math> f=x^2*\frac{d^2y(x)}{dx^2}+x*\frac{dy(x)}{dx}+(x^2-v^2)*y</math>

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)

<math> f=(1-x^2)*\frac{d^2y(x)}{dx^2}-2*x*\frac{dy(x)}{dx}+l(l+1)*y</math>

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)

<math>\frac{dy(x)}{d^2x}-x*\frac{dy(x)}{dx}+n*y(x)</math>

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

<math>x*\frac{d^3y(x)}{d^3x}+x*\frac{dy(x)}{dx}+y(x)-1</math>

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)




<math>\frac{d^2y(x)}{d^2x}+x^3*\frac{dy(x)}{dx}+y(x)-1</math>

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

<math>\frac{d^2y(x)}{d^2x}+(1-x^2)*\frac{dy(x)}{dx}+x*y-1</math>

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;


<math>\frac{d^2(x)}{d^3x}-cos(x)*y(x)-1</math>

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  /
>