## Equações Diferenciais - Soluções Simbólicas

### Differential Equations of First Order

In a first step we write down the differential equation that we want
to solve.
Note the apostroph in front of the **diff**. It is needed to
avoid the immediate computation of a derivative.

eq:'diff(y,x) = -y;

We obtain the equation in a more conventional writing form:

dy

-- = - y

dx

We use **ode2** to compute a solution:

ode2(eq, y, x);

The solution contains the integration constant **%C**.

- x

y = %c %e

To obtain a solution for a selected initial value we use the
function **ic1**
to define the initial condition. For later use, we assign the solution
to the
variable **sol**:

sol: ic1(%, x= 1, y= 8);

This initial condition selects the solution that passes through
point (1, 8).
We obtain:

1 - x

(%o3) y = 8 %e

You can now draw the solution:

plot2d(rhs (sol),[x,-4,4],[y,-10,10]);

### Differential Equations of Second Order

Like differential equations of first, order, differential equations
of second order
are solved with the function **ode2**. To specify an initial
condition, one uses
the function **ic2**, which specifies a point of the solution and
the tangent to
the solution at that point.

#### Example:

eq: 'diff(y, x, 2) + y = 0;

This is:

2

d y

(%o1) --- + y = 0

2

dx

The general solution is again computed with **ode2**:

sol2: ode2(eq, y, x);

We obtain:

(%o2) y = %k1 sin(x) + %k2 cos(x)

This solution has two parameters. There are two substantially
different
ways to compute values for these parameters:

- For an initial value problem, we obtain the parameters for a
particular solution form the value of that particular solution and
its derivative at a given point.
- For a boundary value problem, we need to know the values of the
searched solution at two different points.

For the initial value problem we use the function **ic2**:

ps: ic2(sol2, x=0, y=2, 'diff(y, x)=1);

y = sin(x) + 2 cos(x)

We check that the initial value is satisfied by this solution:

ev (rhs(ps) , x= 0);

2

We check that derivation of the solution satisfies the condition **'diff(y,
x)=1**:

diff(rhs (ps), x);

cos(x) - 2 sin(x)

ev (%, x = 0);

1

We note that it is not possible to combine these two
expressions into:

ev (diff(rhs(ps), x), x=0);

For the boundary value problem we use the function **bc2**:

bc2(sol2, x=0, y=2, x=2, y = -1);

(2 cos(2) + 1) sin(x)

y = 2 cos(x) - ---------------------

sin(2)

It is easy to verify that both boundary conditions are satisfied by
this solution: