Exact Solutions

The statement of some problems requires more than one equation. You enter systems with initial conditions, systems of differential equations, boundary-value problems, or a mixture of these problems using n×1 matrices, where n is the number of equations and conditions involved.


You can also enter such systems in a multiline display after clicking itbpF0.3009in0.3009in0.0701indisplay.wmf.

$\blacktriangleright$ To enter and solve a system of differential equations

1.
Click itbpF0.3009in0.3009in0.0701inmatrix.wmf.

2.
Select 1 column, and set the number of rows equal to the number of equations.

2.
Choose OK.

3.
From the View menu, choose Matrix Lines and Input Boxes (unless Matrix Lines and Input Boxes are already visible) to show where to enter the required equations, and enter the equations.

4.
Leave the insertion point in the matrix.

5.
From the Solve ODE submenu, choose Exact or Laplace.


$\blacktriangleright$ Solve ODE + Exact, Expand

y + y = x
y(0) = 1
(Specify x), Exact solution is : y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = ${\frac{{xe^{x}-e^{x}+2}}{{e^{x}}}}$ = x - 1 + ${\frac{{2}}{{e^{x}}}}$

To solve the second-order initial-value problem y′′ + y = x2, y(0) = 1, y(0) = 1, enter these three equations into a 3×1 matrix and choose Laplace from the Solve ODE submenu.


$\blacktriangleright$ Solve ODE + Laplace

y′′ + y = x2
y(0) = 1   
y(0) = 1   
(Specify x), Laplace solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = x2 -2 + sin x + 3 cos x


The following examples illustrate some of the variety of acceptable notation for entering and solving systems of differential equations in Scientific Notebook.


$\blacktriangleright$ Solve ODE + Laplace

${\dfrac{{dy}}{{dx}}}$ = sin x 
y(0) = 1      
 6pt, Laplace solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = 2 - cos x

Dxxy - y = 0
y(0) = 1
y$\left(\vphantom{ 0}\right.$ 0$\left.\vphantom{ 0}\right)$ = 0
 6pt, Laplace solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = ${\dfrac{{1}}{{2}}}$ex + ${\dfrac{{1}}{{2}}}$e-x  

Dxxxy - y = 0
y(0) = 1
y$\left(\vphantom{ 0}\right.$ 0$\left.\vphantom{ 0}\right)$ = 0
y′′(0) = 0
, Laplace solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = ${\dfrac{{1}}{{3}}}$ex + ${\dfrac{{2}}{{3}}}$e-$\scriptstyle {\frac{{1}}{{2}}}$xcos${\dfrac{{1}}{{2}}}$$\sqrt{{3}}$x


Scientific Notebook introduces a new independent variable in certain instances where none is provided.


$\blacktriangleright$ Solve ODE + Exact

y = x
x = - y
(Specify t), Exact solution is:
x$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = C1sin t + C2cos t
y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = C2sin t - C1cos t


Notice that an exact solution to this problem involves a two-parameter family of solutions..

$\blacktriangleright$ Solve ODE + Laplace

y = x
x = - y
x(0) = 0
y(0) = 1
(Specify t), Laplace solution is:
x$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = - sin t
y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = cos t


Subscripted dependent variables are allowed.


$\blacktriangleright$ Solve ODE + Laplace

Dxy1 + y1 = e2x
y1$\left(\vphantom{ 0}\right.$ 0$\left.\vphantom{ 0}\right)$ = 1
, Laplace solution is: y1$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = ${\dfrac{{2}}{{3}}}$e-x + ${\dfrac{{1}}{{3}}}$e2x

 

Dxxy1 - y1 = 0
y1(0) = 1
y1$\left(\vphantom{ 0}\right.$ 0$\left.\vphantom{ 0}\right)$ = 0
, Laplace solution is: y1$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = ${\dfrac{{1}}{{2}}}$ex + ${\dfrac{{1}}{{2}%
}}$e-x


The next two examples show solutions using Exact for nonlinear equations. Laplace produces no result for these equations, as Laplace transforms are appropriate for linear equations only. Series would also fail in the second example, because ln x does not have a series expansion about x = 0 in powers of x.


$\blacktriangleright$ Solve ODE + Exact

y = y2 + 4
y(0) = - 2
(Specify t), Exact solution is: y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = 2 tan$\left(\vphantom{ 2t-%
\dfrac{1}{4}\pi }\right.$2t - ${\dfrac{{1}}{{4}}}$π$\left.\vphantom{ 2t-%
\dfrac{1}{4}\pi }\right)$

 

(x + 1)y + y = ln x
y(1) = 10
(Specify x), Exact solution is:

y$\displaystyle \left(\vphantom{ x}\right.$x$\displaystyle \left.\vphantom{ x}\right)$ = $\displaystyle {\dfrac{{1}}{{x+1}}}$x ln x - $\displaystyle {\dfrac{{1}}{{x+1}}}$x + $\displaystyle {\dfrac{{21}}{{x+1}}}$ = $\displaystyle {\frac{{%
x\ln x-x+21}}{{x+1}}}$