Editing Initial value problem (section)

==Examples==
A simple example is to solve <math>y'(t) = 0.85 y(t)</math> and <math>y(0) = 19</math>.  We are trying to find a formula for <math>y(t)</math> that satisfies these two equations.

Rearrange the equation so that <math>y</math> is on the left hand side

: <math>\frac{y'(t)}{y(t)} = 0.85</math>

Now integrate both sides with respect to <math>t</math> (this introduces an unknown constant <math>B</math>).

: <math>\int \frac{y'(t)}{y(t)}\,dt = \int 0.85\,dt</math>
: <math>\ln |y(t)| = 0.85t + B </math>

Eliminate the logarithm with exponentiation on both sides

: <math> | y(t) | = e^Be^{0.85t} </math>

Let <math>C</math> be a new unknown constant, <math>C = \pm e^B</math>, so

: <math> y(t) = Ce^{0.85t} </math>

Now we need to find a value for <math>C</math>.  Use <math>y(0) = 19</math> as given at the start and substitute 0 for <math>t</math> and 19 for <math>y</math>

: <math> 19 = C e^{0.85 \cdot 0}</math>
: <math> C = 19 </math>

this gives the final solution of <math> y(t) = 19e^{0.85t}</math>.

;Second example
The solution of

: <math>y'+3y=6t+5,\qquad y(0)=3</math>

can be found to be

: <math>y(t)=2e^{-3t}+2t+1. \,</math>

Indeed,

: <math> \begin{align}
y'+3y &= \tfrac{d}{dt} (2e^{-3t}+2t+1)+3(2e^{-3t}+2t+1) \\
      &= (-6e^{-3t}+2)+(6e^{-3t}+6t+3) \\
      &= 6t+5.
\end{align} </math>
:
'''Third example'''

The solution of

<math>y'=y^{\frac 2 3},\qquad y(0)=0
</math>


<math>\int \frac{y'}{y^{\frac 2 3}}\,dt = \int y^{-\frac 2 3}\,dy  =\int 1\,dt</math>

<math>3 (y(t))^{\frac 1 3}=t+B
</math>

Applying initial conditions we get <math> B=0 </math>, hence the solution:

<math>y(t)= \frac {t^3} {27}
</math>. 


However, the following function is also a solution of the initial value problem:

<math>f(t) = \left\{ \begin{array}{lll} \frac{(t-t_1)^3}{27} & \text{if} & t \leq t_1 \\ 0    & \text{if} & t_1 \leq x \leq t_2 \\ \frac{(t-t_2)^3}{27} & \text{if} & t_2 \leq t \\ \end{array} \right.</math>

The function is differentiable everywhere and continuous, while satisfying the differential equation as well as the initial value problem. Thus, this is an example of such a problem with infinite number of solutions.