The solution of the differential equation \[ \frac{d^{2} y}{d t^{2}}-y=0 \text {, } \] subject to the boundary conditions \( y(0)=1 \) and \( y(\infty)=0 \), is

To solve the differential equation \[ \frac{d^{2} y}{d t^{2}} - y = 0 \] with the boundary conditions \( y(0) = 1 \) and \( y(\infty) = 0 \), follow these steps: 1. **Find the General Solution:** The characteristic equation for the differential equation is: \[ r^2 - 1 = 0 \implies r = \pm 1 \] Therefore, the general solution is: \[ y(t) = A e^{t} + B e^{-t} \] 2. **Apply the Boundary Conditions:** - **First Boundary Condition (\( y(0) = 1 \)):** Substitute \( t = 0 \) into the general solution: \[ y(0) = A e^{0} + B e^{0} = A + B = 1 \quad \Rightarrow \quad A + B = 1 \quad \text{(1)} \] - **Second Boundary Condition (\( y(\infty) = 0 \)):** Examine the behavior of the solution as \( t \to \infty \): \[ y(t) = A e^{t} + B e^{-t} \] For \( y(t) \) to approach 0 as \( t \to \infty \), the term \( A e^{t} \) must not diverge to infinity. This requires \( A = 0 \). Substituting \( A = 0 \) into equation (1): \[ 0 + B = 1 \implies B = 1 \] 3. **Final Solution:** Substituting \( A = 0 \) and \( B = 1 \) back into the general solution: \[ y(t) = e^{-t} \] **Answer:** Its solution is the exponential function y (t ) = e⁻ᵗ