Question

Solve the initial value problem
y′′−2y′+y=0; y(1)=y′(1)=0

A. y(0) = 3
B. y(0) = 1
C. y(0) = 2
D. y(0) = 0

Answer 1

Answer on Question #82234 – Math – Differential Equations

Question

Solve the initial value problem

$$y'' - 2y' + y = 0; \, y(1) = y'(1) = 0$$

A. $y(0) = 3$

B. $y(0) = 1$

C. $y(0) = 2$

D. $y(0) = 0$

Solution

Let us construct the characteristic equation:

$$\begin{array}{l} k^2 - 2k + 1 = 0 \\ (k - 1)^2 = 0 \end{array}$$

This equation has two equal roots $k = 1$;

The solution can be written in the form:

$$y(x) = e^{kx}(C_1 + C_2x)$$ $$y(x) = e^x(C_1 + C_2x)$$

Then we substitute the initial conditions:

$$\begin{array}{l} y(1) = e(C_1 + C_2) = 0 \\ y'(1) = (C_1 e^x + C_2 x e^x + C_2 e^x) \big|_{x=1} = e(C_1 + 2C_2) = 0 \end{array}$$

We obtain the system:

$$\left\{ \begin{array}{l} C_1 + C_2 = 0 \\ C_1 + 2C_2 = 0 \end{array} \right.$$

Therefore, $C_1 = 0$, $C_2 = 0$.

Only the zero solution satisfies the initial conditions: $y \equiv 0$

Thus, $y(0) = 0$, and the right answer is D).

Answer: D) $y(0) = 0$.

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