Question

Consider the initial value problem
\(\frac{dy}{dx}+αy=0, \\ y(0)=1,\)
where α ∈ \(\R\). Then

• A. there is an α such that y(1) = 0
• B. there is a unique α such that \(\lim\limits_{x\rightarrow \infin}y(x)=0\)
• C. there is NO α such that y(2) = 1
• D. there is a unique α such that y(1) = 2

Answer 1

The Correct Option is D

We have the initial value problem given by the differential equation: 

\(\frac{dy}{dx} + \alpha y = 0\), \(y(0) = 1\)

To solve this differential equation, recognize that it is a first-order linear homogeneous differential equation which can be solved using separation of variables.

Rearrange the equation:

\(\frac{dy}{y} = -\alpha \, dx\)

Integrating both sides gives:

\(\int \frac{dy}{y} = \int -\alpha \, dx\)

\(\ln |y| = -\alpha x + C\)

Exponentiating both sides yields:

\(y = e^{-\alpha x + C}\)

Let \(e^C = C_1\), so:

\(y = C_1 e^{-\alpha x}\)

Using the initial condition \(y(0) = 1\):

\(1 = C_1 e^{0} = C_1\)

This gives \(C_1 = 1\). Therefore, the solution is:

\(y = e^{-\alpha x}\)

We now analyze each option with this solution:

• For y(1) = 2: Set \(e^{-\alpha} = 2\). This implies \(-\alpha = \ln 2\), hence \(\alpha = -\ln 2\), which is unique.
• For y(1) = 0: Since \(e^{-\alpha x}\) is never zero for real \(\alpha\), this is impossible.
• For y(2) = 1: This would imply \(e^{-2\alpha} = 1\). This occurs for \(\alpha = 0\), contradicting "NO \(\alpha\)" condition.
• For \(\lim\limits_{x\rightarrow \infin}y(x)=0\): For any \(\alpha > 0\), \(y\) approaches zero. Thus, there is not a unique \(\alpha\).

Therefore, the correct answer is that there is a unique \(\alpha\) such that \(y(1) = 2\).