Solving yy 0 with Boundary Conditions y(-1) 0 and y(1) 0

To solve the differential equation yy 0 with the boundary conditions y(-1) 0 and y(1) 0, we'll follow a systematic approach. This method involves solving the differential equation, applying the boundary conditions, and determining the constants accordingly.

Step 1: Solve the Homogeneous Equation

Let's start by solving the homogeneous equation. Substituting y e^{rx} into the differential equation, we get:

y y 0 translates to r^2 e^{rx} - e^{rx} 0, which simplifies to:

r^2 - 1 0

The characteristic equation yields the roots:

r i and r -i

Therefore, the general solution to the differential equation is:

y(x) C_1 cos(x) C_2 sin(x)

Here, C_1 and C_2 are constants that need to be determined using the boundary conditions.

Step 2: Apply Boundary Conditions

Now we'll apply the boundary conditions one by one.

First Boundary Condition: y(-1) 0

Substituting x -1 into the general solution:

y(-1) C_1 cos(-1) C_2 sin(-1) C_1 cos(1) - C_2 sin(1)

This can be rearranged to:

C_1 cos(1) C_2 sin(1)

Second Boundary Condition: y(1) 0

Substituting x 1 into the general solution:

y(1) C_1 cos(1) C_2 sin(1)

This can also be rearranged to:

C_1 cos(1) C_2 sin(1) 0

Step 3: Solve the System of Equations

Now we have two equations:

C_1 cos(1) C_2 sin(1) C_1 cos(1) C_2 sin(1) 0

From the first equation, we can express C_2 in terms of C_1 as:

C_2 C_1 cos(1) / sin(1)

Substituting this into the second equation:

C_1 cos(1) (C_1 cos(1) / sin(1)) sin(1) 0

This simplifies to:

2C_1 cos(1) 0

This gives us two cases:

C_1 0 cos(1) 0

cos(1) 0 is not true since 1 is not an odd multiple of π/2.

Step 4: Determine Constants

If C_1 0, substituting this back into the first equation:

C_1 cos(1) C_2 sin(1) 0

This implies C_2 0 since neither sin(1) nor C_1 are zero.

Therefore, only the trivial solution C_1 0 and C_2 0 satisfies both boundary conditions.

Conclusion: The only solution that satisfies both boundary conditions is:

y(x) 0

This means the trivial solution is the only solution to the boundary value problem given the conditions.