Answer:An initial condition is an extra bit of information about a differential equation that tells you the value of the function at a particular point. Differential equations with initial conditions are commonly called initial value problems.
The video above uses the example
{
d
y
d
x
=
cos
(
x
)
y
(
0
)
=
−
1
to illustrate a simple initial value problem. Solving the differential equation without the initial condition gives you
y
=
sin
(
x
)
+
C
.
Once you get the general solution, you can use the initial value to find a particular solution which satisfies the problem. In this case, plugging in
0
for
x
and
−
1
for
y
gives us
−
1
=
C
, meaning that the particular solution must be
y
=
sin
(
x
)
−
1
.
So the general way to solve initial value problems is: - First, find the general solution while ignoring the initial condition. - Then, use the initial condition to plug in values and find a particular solution.
Two additional things to keep in mind: First, the initial value doesn't necessarily have to just be
y
-values. Higher-order equations might have an initial value for both
y
and
y
′
, for example.
Second, an initial value problem doesn't always have a unique solution. It's possible for an initial value problem to have multiple solutions, or even no solution at all.
Explanation:
Answer:
a tactile input device
Explanation:
Based on the scenario being described it can be said that the best assistive technology for Bernard would be a tactile input device. Such a device would allow Bernard to physically interact with the print on the chalkboard or overhead screen from the device itself without having to actually get close to the chalkboard or screen. Thus allowing him to view its content from the palm of his hand.
Windows Movie Maker, iMovie, Final Cut Pro (x), QuickTime?
Answer:
showProduct(int,double)
for example: showProduct(10,10.5) is the correct answer even showProduct(10,10.0) is also correct but showProduct(10.0,10.5) or showProduct(10,10) or showProduct(10.0,10) are wrong calls.
Explanation:
The code is
- <em>public static void showProduct (int num1, double num2){</em>
- <em> int product;</em>
- <em> product = num1*(int)num2;</em>
- <em> System.out.println("The product is "+product);</em>
- <em> }</em>
showProduct is function which asks for two arguments whenever it is called, first one is integer and second one is of type double which is nothing but decimal point numbers. Generally, in programming languages, 10 is treated as integer but 10.0 is treated as decimal point number, but in real life they are same.
If showProduct( 10,10.0) is called the output will be 'The product is 100'.
Strange fact is that, if you enter showProduct(10,10.5) the output will remain same as 'The product is 100'. This happens because in the 3rd line of code,which is <em>product=num1*(int)num2</em>, (int) is placed before num2 which makes num2 as of type integer, which means whatever the value of num2 two is given, numbers after decimal is erased and only the integer part is used there.
This is necessary in JAVA and many other programming languages as you <u>cannot</u><u> multiply two different datatypes</u> (here one is int and another is double). Either both of them should be of type int or both should be of type double.
