Question

The function \( y(t) \) satisfies \[ t^2 y''(t) - 2t y'(t) + 2y(t) = 0, \] where \( y'(t) \) and \( y''(t) \) denote the first and second derivatives of \( y(t) \), respectively. Given \( y'(0) = 1 \) and \( y'(1) = -1 \), the maximum value of \( y(t) \) over \( [0, 1] \) is _________ (rounded off to two decimal places).

Hint:

For Cauchy-Euler equations, assume a solution of the form \( y(t) = t^r \). After solving the characteristic equation, apply the initial conditions to find the constants and determine the maximum value of the function.

Answer 1

Correct Answer: 0.25

Step 1: Substitute the assumed solution \( y(t) = t^r \):
First, compute the first and second derivatives of \( y(t) \): \[ y'(t) = r t^{r-1}, \quad y''(t) = r(r-1) t^{r-2}. \] Now, substitute these into the differential equation: \[ t^2 (r(r-1) t^{r-2}) - 2t (r t^{r-1}) + 2t^r = 0. \] Simplify: \[ r(r-1) t^r - 2r t^r + 2t^r = 0. \] Factor out \( t^r \) (since \( t \neq 0 \)): \[ t^r \left[ r(r-1) - 2r + 2 \right] = 0. \] Simplifying the expression inside the brackets: \[ r(r-1) - 2r + 2 = r^2 - r - 2r + 2 = r^2 - 3r + 2. \] Thus, the characteristic equation is: \[ r^2 - 3r + 2 = 0. \] Step 2: Solve the characteristic equation:
Factor the quadratic equation:
\[ (r - 1)(r - 2) = 0. \] Therefore, the solutions are \( r = 1 \) and \( r = 2 \). Step 3: General solution:
The general solution to the differential equation is:
\[ y(t) = C_1 t + C_2 t^2, \] where \( C_1 \) and \( C_2 \) are constants to be determined from the initial conditions. Step 4: Apply the initial conditions:
We are given \( y'(0) = 1 \) and \( y'(1) = -1 \).
First, compute \( y'(t) \): \[ y'(t) = C_1 + 2C_2 t. \] Apply \( y'(0) = 1 \): \[ C_1 + 2C_2 \cdot 0 = 1 \quad \Rightarrow \quad C_1 = 1. \] Apply \( y'(1) = -1 \): \[ C_1 + 2C_2 \cdot 1 = -1 \quad \Rightarrow \quad 1 + 2C_2 = -1 \quad \Rightarrow \quad 2C_2 = -2 \quad \Rightarrow \quad C_2 = -1. \] Step 5: Final solution: - Thus, the solution for \( y(t) \) is: \[ y(t) = t - t^2. \] Step 6: Find the maximum value of \( y(t) \) over \( [0, 1] \):
To find the maximum value of \( y(t) \), take the derivative: \[ y'(t) = 1 - 2t. \] Set \( y'(t) = 0 \) to find the critical points: \[ 1 - 2t = 0 \quad \Rightarrow \quad t = \frac{1}{2}. \] Evaluate \( y(t) \) at \( t = \frac{1}{2} \): \[ y\left( \frac{1}{2} \right) = \frac{1}{2} - \left( \frac{1}{2} \right)^2 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}. \] Evaluate \( y(t) \) at the endpoints:
\( y(0) = 0 - 0 = 0 \),
\( y(1) = 1 - 1^2 = 0 \).
The maximum value of \( y(t) \) on \( [0, 1] \) is \( \frac{1}{4} \).
Thus, the maximum value of \( y(t) \) over \( [0, 1] \) is \( \frac{1}{4} \), or 0.25.