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Finding c1 and c2 to Satisfy Given Conditions
As a high school mathematics teacher, I often encounter questions where students need to find the values of the constants c1 and c2 in the function y(x) = c1 sin(x) + c2 cos(x) to satisfy certain conditions. In this article, I will guide you through the step-by-step process of solving two such problems.
Problem 1: Find c1 and c2 such that y(0) = 0 and y'(π/2) = 1
To solve this problem, we need to use the given conditions to set up a system of equations and solve for c1 and c2.
- Condition 1: y(0) = 0
Substituting x = 0 into the function, we get:
y(0) = c1 sin(0) + c2 cos(0)
y(0) = c1 × 0 + c2 × 1
y(0) = c2
Therefore, c2 = 0.
- Condition 2: y'(π/2) = 1
To find the derivative of y(x), we use the formula:
y'(x) = c1 cos(x) - c2 sin(x)
Substituting x = π/2, we get:
y'(π/2) = c1 cos(π/2) - c2 sin(π/2)
y'(π/2) = c1 × 0 - c2 × 1
y'(π/2) = -c2
Therefore, c2 = -1.
Substituting the values of c2 into the original function, we get: y(x) = c1 sin(x) + 0 cos(x) y(x) = c1 sin(x)
To find the value of c1, we can use either of the given conditions. Let's use the first condition: y(0) = c1 sin(0) = 0 c1 = 0
Therefore, the final solution is: c1 = 0 c2 = -1 y(x) = 0 sin(x) - 1 cos(x) = -cos(x)
Problem 2: Find c1 and c2 such that y(0) = 1 and y'(π) = 1
- Condition 1: y(0) = 1
Substituting x = 0 into the function, we get:
y(0) = c1 sin(0) + c2 cos(0)
y(0) = c1 × 0 + c2 × 1
y(0) = c2
Therefore, c2 = 1.
- Condition 2: y'(π) = 1
To find the derivative of y(x), we use the formula:
y'(x) = c1 cos(x) - c2 sin(x)
Substituting x = π, we get:
y'(π) = c1 cos(π) - c2 sin(π)
y'(π) = c1 × (-1) - c2 × 0
y'(π) = -c1
Therefore, c1 = -1.
Substituting the values of c1 and c2 into the original function, we get: y(x) = -1 sin(x) + 1 cos(x)
Therefore, the final solution is: c1 = -1 c2 = 1 y(x) = -1 sin(x) + 1 cos(x)
I hope this article has helped you understand the process of finding the values of c1 and c2 to satisfy given conditions for the function y(x) = c1 sin(x) + c2 cos(x). If you have any further questions, feel free to ask.