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:
12. for (i = 0 ; i < testGrades.length ; i+=1 ){
13. if (testGrades[i] > 100){
14. sumExtra = sumExtra + testGrades[i] - 100;}
15. }
Explanation:
We first iterate through the entire testGrades array. For each test score that is in testGrades ( that is testGrades[i] ), we see whether or not the test grade is above 100 (See line 12) . If test grade is greater than 100, this means we have extra credit. We simply subtract 100 from the test grade, add it with the previous value of sumExtra and store the value back in sumExtra(see line 14). Once i is greater than the length of the test grades, the loop is exited. We can now print sumExtra to obtain the result.
Answer is: Yes
<u>Explanation:</u>
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