We are given the second-order linear homogeneous differential equation $y'' + 4y = 0$ with boundary conditions $y(0) = 1$ and $y(\frac{\pi}{4}) = 2$. We are asked to find the value of $y'(0)$.

Applied MathematicsDifferential EquationsSecond-Order Differential EquationsBoundary Value ProblemTrigonometry
2025/5/14

1. Problem Description

We are given the second-order linear homogeneous differential equation y+4y=0y'' + 4y = 0 with boundary conditions y(0)=1y(0) = 1 and y(π4)=2y(\frac{\pi}{4}) = 2. We are asked to find the value of y(0)y'(0).

2. Solution Steps

The characteristic equation for the differential equation y+4y=0y'' + 4y = 0 is r2+4=0r^2 + 4 = 0.
Solving for rr, we get r2=4r^2 = -4, so r=±2ir = \pm 2i.
Since the roots are complex conjugates, the general solution is of the form
y(x)=c1cos(2x)+c2sin(2x)y(x) = c_1 \cos(2x) + c_2 \sin(2x).
We are given the boundary condition y(0)=1y(0) = 1.
Substituting x=0x = 0 into the general solution, we get
y(0)=c1cos(0)+c2sin(0)=c1(1)+c2(0)=c1y(0) = c_1 \cos(0) + c_2 \sin(0) = c_1(1) + c_2(0) = c_1.
Since y(0)=1y(0) = 1, we have c1=1c_1 = 1.
Now the solution is y(x)=cos(2x)+c2sin(2x)y(x) = \cos(2x) + c_2 \sin(2x).
We are given the boundary condition y(π4)=2y(\frac{\pi}{4}) = 2.
Substituting x=π4x = \frac{\pi}{4} into the solution, we get
y(π4)=cos(2π4)+c2sin(2π4)=cos(π2)+c2sin(π2)=0+c2(1)=c2y(\frac{\pi}{4}) = \cos(2 \cdot \frac{\pi}{4}) + c_2 \sin(2 \cdot \frac{\pi}{4}) = \cos(\frac{\pi}{2}) + c_2 \sin(\frac{\pi}{2}) = 0 + c_2(1) = c_2.
Since y(π4)=2y(\frac{\pi}{4}) = 2, we have c2=2c_2 = 2.
So the solution is y(x)=cos(2x)+2sin(2x)y(x) = \cos(2x) + 2 \sin(2x).
Now we need to find y(x)y'(x).
y(x)=2sin(2x)+4cos(2x)y'(x) = -2 \sin(2x) + 4 \cos(2x).
We want to find y(0)y'(0).
y(0)=2sin(0)+4cos(0)=2(0)+4(1)=0+4=4y'(0) = -2 \sin(0) + 4 \cos(0) = -2(0) + 4(1) = 0 + 4 = 4.

3. Final Answer

y(0)=4y'(0) = 4

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