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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:
import java.util.Scanner;
class Main {
public static int calcSeries(int n) {
int sum = 0;
for(int i=10; i>=n; i--) {
sum += i;
}
return sum;
}
public static void main(String[] args) {
Scanner reader = new Scanner(System.in);
int n = 0;
do {
System.out.print("Enter n: ");
n = reader.nextInt();
if (n >= 10) {
System.out.println("Please enter a value lower than 10.");
}
} while (n >= 10);
reader.close();
System.out.printf("sum: %d\n", calcSeries(n));
}
}
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