DSolve(equation, dependent(independent), values, mode=1)

DSolve is a differential equation solver.

The derivative of the variable is notated using derivative notation. SpaceTime uses the single quotation character to represent the derivative of a variable or function.

f'(x) - First derivative
f''(x) - Second derivative
f'''(x) - Third derivative

Derivative Examples
Solve: y' + y = sin(t)
DSolve(y'(t)+y(t)=sin(t), y(t), .....)

Solve: y'' + 2y' + y = x
DSolve(y''(x)+2y'(x)+y(x)=x, y(x), .....)

The definite integral of the variable is represented by I(f(t), t) and denotes the integral of the variable f(t) with lower bound 0 and upper bound t. These bounds cannot be changed.

Integral Examples
∫(y) + y = sin(t)
DSolve( I(y(t),t) + y(t) = sin(t), y(t), .....)

The upper case character I indicates the integral and is a mandatory condition.

Initial values
When no initial values are given, then simple write the word no. The result returned contains the Constant C@1.

For first order differential equations the initial value can be given as y(t)=value. When you ommit y(t), then y(0) is assumed.

For higher order differential equations, the initial values should be given as a list and are always referred to zero : [y(0), y'(0), y' '(0), ...]

Examples with different initial values
Solve: y' + y = sin(t) with y(0) = 5
DSolve(y'(t) + y(t) = sin(t) , y(t) , y(0) = 5)
or
DSolve(y(t)' + y(t) = sin(t) , y(t) , 5)

Solve: y' + y = sin(t) with y(2) = 5
DSolve(y(t)' + y(t) = sin(t) , y(t) , y(2)=5)

Solve: y'' + 2y' + y = x with y(0) = 1 and y'(0) = 2
DSolve(y''(x) + 2y'(x) + y(x) = x, y(x), [1,2])

Solve: y'' + 2y' + y = x with no initial values
DSolve(y''(x) + 2y'(x) + y(x) = x, y(x), no)

Different form for exact first order differential equations.
Exact first order differential equation can be given with d@x , d@t , d@y, ect.

First Order Examples
Solve: dy = sin(t) * dt
DSolve(d@y = sin(t) * d@t, y(t), .....)

Of course, this can also be written as the following equation.
DSolve(y'(t), t) = sin(t) , y(t), ...)
Note: The variable to be solved should still be written as y(t).

Results
Normally, DSolve tries to return the result as y(t) = function(t).

In case y(t) cannot be represented in a simple form then the result is returned in implicit form.
function(y , t) = 0

In the latter case, the text Implicit result is added to call the users attention that the result is not of the form y(t)=function(t).

Examples

DSolve(y''(x)+2y'(x)+y(x)=x, y(x), no)

DSolve(y''(x)+2y'(x)+y(x)=x, y(x), [1, 5])

DSolve(R*q'(t)+q(t)/C=U*sin(t), q(t), no)

DSolve(4*q'(t)+2q(t)=U*sin(t), q(t), y(0)=0)

DSolve(exp(x)*sin(y(x))*d@x+((y(x)+exp(x)*cos(y(x)))*d@y, y(x), no)

DSolve(y'(t)=3t^2*exp(-y(t)), y(t), y(0)=1)

DSolve(x^2*y'(x)+3*x*y(x)=x^3, y(x), y(1)=A)

DSolve(x'(t)=6*sin(t)/x(t), x(t), y(\pi/4)=2)

DSolve(x'(t)=6*sin(t)/x(t), x(t), y(\pi/4)=2, 0)

DSolve(f(z)^2+2*z*f(z)*f'(z), f(z), no, 0)

DSolve(f(z)^2+2*z*f(z)*f'(z), f(z), no)

DSolve(f(z)^2+2*z*f(z)*f'(z), f(z), f(1)=5)

DSolve((a+b*x)+(e+x+c)*y'(x), y(x), no)

DSolve(x*f'(x)+x+f(x), f(x), f(1)=a, 0)

DSolve(y'(t)+2a*y(t)=t^2, y(t), y(1)=0)

References

http://en.wikipedia.org/wiki/Ordinary_differential_equation

Related Functions
D, Diff, Integrate, Solve