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In this problem, y=c1cos(3x)+c2sin(3x)y = c_1 \cos(3x) + c_2 \sin(3x) is a two-parameter family of solutions of the second-order DE y+9y=0y'' + 9y = 0. If possible, find a solution of the differential equation that satisfies the given side conditions. The conditions specified at two different points are called boundary conditions. (If not possible, enter NOT.)

Given: y(0)=0y(0) = 0, y(π6)=8y\left(\frac{\pi}{6}\right) = 8

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Question :

In this problem, y=c1cos(3x)+c2sin(3x)y = c_1 \cos(3x) + c_2 \sin(3x) is a two-parameter family of solutions of the second-order de y+9y=0y'' + 9y = 0. if possible, find a solution of the differential equation that satisfies the given side conditions. the conditions specified at two different points are called boundary conditions. (if not possible, enter not.)

given: y(0)=0y(0) = 0, y(π6)=8y\left(\frac{\pi}{6}\right) = 8

In this problem, y = c_1 \cos(3x) + c_2 \sin(3x) is a two-parameter family of  | Doubtlet.com

Solution:

Neetesh Kumar

Neetesh Kumar | November 09, 2024

Differential Equation Homework Help

This is the solution to Math 2A, section 13Z, Fall 2023 | WebAssign
Math002ACh1Sec02 (Homework) Question - 10
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Step-by-Step-Solution:

Step 1: Write Down the General Solution and Apply y(0)=0y(0) = 0

The general solution to the differential equation y+9y=0y'' + 9y = 0 is: y=c1cos(3x)+c2sin(3x)y = c_1 \cos(3x) + c_2 \sin(3x)

We are given the initial condition y(0)=0y(0) = 0. Substitute x=0x = 0 into the general solution: y(0)=c1cos(30)+c2sin(30)y(0) = c_1 \cos(3 \cdot 0) + c_2 \sin(3 \cdot 0) y(0)=c11+c20y(0) = c_1 \cdot 1 + c_2 \cdot 0 y(0)=c1y(0) = c_1

Since y(0)=0y(0) = 0, we have: c1=0c_1 = 0

Step 2: Substitute c1=0c_1 = 0 into the Solution

With c1=0c_1 = 0, the general solution simplifies to: y=c2sin(3x)y = c_2 \sin(3x)

Step 3: Apply the Boundary Condition y(π6)=8y\left(\frac{\pi}{6}\right) = 8

Now we use the boundary condition y(π6)=8y\left(\frac{\pi}{6}\right) = 8.

Substitute x=π6x = \frac{\pi}{6} into the simplified solution: y(π6)=c2sin(3π6)y\left(\frac{\pi}{6}\right) = c_2 \sin\left(3 \cdot \frac{\pi}{6}\right) y(π6)=c2sin(π2)y\left(\frac{\pi}{6}\right) = c_2 \sin\left(\frac{\pi}{2}\right)

Since sin(π2)=1\sin\left(\frac{\pi}{2}\right) = 1, this becomes: 8=c218 = c_2 \cdot 1 c2=8c_2 = 8

Step 4: Write the Particular Solution

With c1=0c_1 = 0 and c2=8c_2 = 8, the particular solution to the differential equation is: y=8sin(3x)y = 8 \sin(3x)

Final Answer:

The solution that satisfies the boundary conditions is:

y=8sin(3x)y = \boxed{8 \sin(3x)}



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