Solving differential equations : desolve deSolve
dsolve

desolve (or deSolve) can solve :

• linear differential equations with constant coefficients,
• first order linear differential equations,
• first order differential equations without y,
• first order differential equations without x,
• first order differential equations with separated variables,
• first order homogeneous differential equations (y' = F(y/x)),
• first order differential equations with integrating factor,
• first order Bernoulli differential equations ( a(x)y' + b(x)y = c(x)yⁿ),
• first order Clairaut differential equations ( y = x*y' + f (y')).

desolve takes as arguments :

• if the independant variable is the current variable (here supposed to be x),
  • the differential equation (or the list of the differential equation and of the initial conditions)
  • the unknown (usually y).
  In the differential equation, the function y is denoted by y, it's first derivative y$\prime$ is denoted by $\tt y'$, and it's second derivative y'' is written $\tt y''$.
  For example desolve(y''+2*y'+y,y) or
  desolve([y''+2*y'+y,y(0)=1,y'(0)=0],y).
• if the independant variable is not the current variable, for example t instead of x,
  • the differential equation (or the list of the differential equation and of the initial conditions),
  • the variable, e.g. t
  • the unknown as a variable y or as a function y(t).
  In the differential equation, the function y is denoted by y(t), it's derivative y$\prime$ is denoted by diff(y(t),t), and it's second derivative y'' is denoted by diff(y(t),t$2).
  For example :
  desolve(diff(y(t),t$2)+2*diff(y(t),t)+y(t),y(t)); or
  desolve(diff(y(t),t$2)+2*diff(y(t),t)+y(t),t,y); and
  

  desolve([diff(y(t),t$2)+2*diff(y(t),t)+y(t),
           y(0)=1,y'(0)=0],y(t)); or
  desolve([diff(y(t),t$2)+2*diff(y(t),t)+y(t), 
           y(0)=1,y'(0)=0],t,y);
  

If there is no initial conditions (or one initial condition for a second order equation), desolve returns the general solution in terms of constants of integration c_0, c_1, where y(0)=c_0 and y'(0)=c_1, or a list of solutions.
Examples

• Examples of second linear differential equations with constant coefficients.
  1. Solve :
     y'' + y = cos(x)
     Input (typing twice prime for y''):
     desolve(y''+y=cos(x),y)
     or input :
     desolve((diff(diff(y))+y)=(cos(x)),y)
     Output :
     c_0*cos(x)+(x+2*c_1)*sin(x)/2
     c_0, c_1 are the constants of integration : y(0)=c_0 and y'(0)=c_1.
     If the variable is not x but t, input :
     desolve(derive(derive(y(t),t),t)+y(t)=cos(t),t,y)
     Output :
     c_0*cos(t)+(t+2*c_1)/2*sin(t)
     c_0, c_1 are the constants of integration : y(0)=c_0 and y'(0)=c_1.
  2. Solve :
     y'' + y = cos(x),    y(0) = 1
     Input :
     desolve([y''+y=cos(x),y(0)=1],y)
     Output :
     [cos(x)+(x+2*c_1)/2*sin(x)]
     the components of this vector are solutions (here there is just one component, so we have just one solution depending of the constant c_1).
  3. Solve :
     y'' + y = cos(x)    (y(0))² = 1
     Input :
     desolve([y''+y=cos(x),y(0)^2=1],y)
     Output :
     [-cos(x)+(x+2*c_1)/2*sin(x),cos(x)+(x+2*c_1)/2*sin(x)]
     each component of this list is a solution, we have two solutions depending on the constant c_1 (y'(0) = c₁) and corresponding to y(0) = 1 and to y(0) = - 1.
  4. Solve :
     y'' + y = cos(x),    (y(0))² = 1    y'(0) = 1
     Input :
     desolve([y''+y=cos(x),y(0)^2=1,y'(0)=1],y)
     Output :
     [-cos(x)+(x+2)/2*sin(x),cos(x)+(x+2)/2*sin(x)]
     each component of this list is a solutions (we have two solutions).
  5. Solve :
     y'' + 2y' + y = 0
     Input :
     desolve(y''+2*y'+y=0,y)
     Output :
     (x*c_0+x*c_1+c_0)*exp(-x)
     the solution depends of 2 constants of integration : c_0, c_1 (y(0)=c_0 and y'(0)=c_1).
  6. Solve :
     y'' - 6y' + 9y = xe^3x
     Input:
     desolve(y''-6*y'+9*y=(x*exp(3*x),y)
     Output :
     (x^3+(-(18*x))*c_0+6*x*c_1+6*c_0)*1/6*exp(3*x)
     the solution depends on 2 constants of integration : c_0, c_1 (y(0)=c_0 and y'(0)=c_1).
• Examples of first order linear differential equations.
  1. Solve :
     xy' + y - 3x² = 0
     Input :
     desolve(x*y'+y-3*x^2,y)
     Output :
     (3*1/3*x^3+c_0)/x
  2. Solve :
     y' + x*y = 0, y(0) = 1
     Input :
     desolve([y'+x*y=0, y(0)=1]),y)
     or :
     desolve((y'+x*y=0) && (y(0)=1),y)
     Output :
     [1/(exp(1/2*x^2))]
  3. Solve :
     x(x² - 1)y' + 2y = 0
     Input :
     desolve(x*(x^2-1)*y'+2*y=0,y)
     Output :
     (c_0)/((x^2-1)/(x^2))
  4. Solve :
     x(x² -1)y' + 2y = x²
     Input :
     desolve(x*(x^2-1)*y'+2*y=x^2,y)
     Output :
     (ln(x)+c_0)/((x^2-1)/(x^2))
  5. If the variable is t instead of x, for example :
     t(t² -1)y'(t) + 2y(t) = t²
     Input :
     desolve(t*(t^2-1)*diff(y(t),t)+2*y(t)=(t^2),y(t))
     Output :
     (ln(t)+c_0)/((t^2-1)/(t^2))
  6. Solve :
     x(x² -1)y' + 2y = x², y(2) = 0
     Input :
     desolve([x*(x^2-1)*y'+2*y=x^2,y(0)=1],y)
     Output :
     [(ln(x)-ln(2))*1/(x^2-1)*x^2]
  7. Solve :
     $$\sqrt{{1+x^2}}$$y' - x - y = $$\sqrt{{1+x^2}}$$
     Input :
     desolve(y'*sqrt(1+x^2)-x-y-sqrt(1+x^2),y)
     Output :
     (-c_0+ln(sqrt(x^2+1)-x))/(x-sqrt(x^2+1))

• Examples of first differential equations with separated variables.
  1. Solve :
     y' = 2$$\sqrt{{y}}$$
     Input :
     desolve(y'=2*sqrt(y),y)
     Output :
     [x^2+-2*x*c_0+c_0^2]
  2. Solve :
     xy'ln(x) - y(3 ln(x) + 1) = 0
     Input :
     desolve(x*y'*ln(x)-(3*ln(x)+1)*y,y)
     Output :
     c_0*x^3*ln(x)

• Examples of Bernoulli differential equations a(x)y' + b(x)y = c(x)yⁿ where n is a real constant.
  The method used is to divide the equation by yⁿ, so that it becomes a first order linear differential equation in u = 1/y^n-1.
  1. Solve :
     xy' + 2y + xy² = 0
     Input :
     desolve(x*y'+2*y+x*y^2,y)
     Output :
     [1/(exp(2*ln(x))*(-1/x+c_0))]
  2. Solve :
     xy' - 2y = xy³
     Input :
     desolve(x*y'-2*y-x*y^3,y)
     Output :
     [((-2*1/5*x^5+c_0)*exp(-(4*log(x))))^(1/-2),
     -((-2*1/5*x^5+c_0)*exp(-(4*log(x))))^(1/-2)]
  3. Solve :
     x²y' - 2y = xe⁽4/x)y³
     Input :
     desolve(x*y'-2*y-x*exp(4/x)*y^3,y)
     Output :
     [((-2*ln(x)+c_0)*exp(-(4*(-(1/x)))))^(1/-2),
     -(((-2*ln(x)+c_0)*exp(-(4*(-(1/x)))))^(1/-2))]

• Examples of first order homogeneous differential equations (y' = F(y/x), the method of integration is to search t = y/x instead of y).
  1. Solve :
     (3x³y' = y(3x² - y²)
     Input :
     desolve(3*x^3*diff(y)=((3*x^2-y^2)*y),y)
     Output :
     [0,pnt[c_0*exp((3*1/2)/(` t`^2)),` t`*c_0*exp((3*1/2)/(` t`^2))]]
     hence the solutions are y = 0 and the familiy of curves of parametric equation x = c₀exp(3/(2t²)), y = t*c₀exp(3/(2t²)) (the parameter is denoted by ` t` in the answer).
  2. Solve :
     xy' = y + $$\sqrt{{x^2+y^2}}$$
     Input :
     desolve(x*y'=y+sqrt(x^2+y^2),y)
     Output :
     [(-i)*x,(i)*x,pnt[c_0/(sqrt(` t`^2+1)-` t`),(` t`*c_0)/(sqrt(` t`^2+1)-` t`)]]
     hence the solutions are :
     y = ix, y = - ix
     and the family of curves of parametric equations
     x = c₀/($$\sqrt{{t^2+1}}$$ - t), y = t*c₀/($$\sqrt{{t^2+1}}$$ - t)
     (the parameter is denoted by ` t` in the answer).

• Examples of first order differential equations with an integrating factor. By multiplying the equation by a function of x, y, it becomes a closed differential form.
  1. Solve :
     yy' + x
     Input :
     desolve(y*y'+x,y)
     Output :
     [sqrt(-2*c_0-x^2),-(sqrt(-2*c_0-x^2))]
     In this example, xdx + ydy is closed, the integrating factor was 1.
  2. Solve :
     2xyy' + x² - y² + a² = 0
     Input :
     desolve(2*x*y*y'+x^2-y^2+a^2,y)
     Output :
     [sqrt(a^2-x^2-c_1*x),-(sqrt(a^2-x^2-c_1*x))]
     In this example, the integrating factor was 1/x².

• Example of first order differential equations without x.
  Solve :
  (y + y')⁴ + y' + 3y = 0
  This kind of equations can not be solved directly by Xcas, we explain how to solve them with it's help. The idea is to find a parametric representation of F(u, v) = 0 where the equation is F(y, y') = 0, Let u = f (t), v = g(t) be such a parametrization of F = 0, then y = f (t) and dy/dx = y' = g(t). Hence
  dy/dt = f'(t) = y'*dx/dt = g(t)*dx/dt
  The solution is the curve of parametric equations x(t), y(t) = f (t), where x(t) is solution of the differential equation g(t)dx = f'(t)dt.
  Back to the example, we put y + y' = t, hence:
  y = - t - 8*t⁴,    y' = dy/dx = 3*t + 8*t⁴    dy/dt = - 1 - 32*t³
  therefore
  (3*t + 8*t⁴)*dx = (- 1 - 32*t³)dt
  Input :
  desolve((3*t+8*t^4)*diff(x(t),t)=(-1-32*t^3),x(t))
  Output :
  -11*1/9*ln(8*t^3+3)+1/-9*ln(t^3)+c_0
  eventually the solution is the curve of parametric equation :
  x(t) = - 11*1/9*ln(8*t³ +3) + 1/ -9*ln(t³) + c₀,    y(t) = - t - 8*t⁴

• Examples of first order Clairaut differential equations ( y = x*y' + f (y')).
  The solutions are the lines D_m of equation y = mx + f (m) where m is a real constant.
  1. Solve :
     xy' + y'³ - y) = 0
     Input :
     desolve(x*y'+y'^3-y),y)
     Output :
     c_0*x+c_0^3
  2. Solve :
     y - xy' = $$\sqrt{{a^2+b^2*y'^2}}$$ = 0
     Input :
     desolve((y-x*y'-sqrt(a^2+b^2*y'^2),y)
     Output :
     c_0*x+sqrt(a^2+b^2*c_0^2)