Consider the initial value problem dy/dx + αy = 0, y(0) = 1, where α ∈ R. Then (a) there is an α such that y(1) = 0.

Understand the Problem

The question presents an initial value problem involving a differential equation and asks whether there exists a value of α such that the solution satisfies y(1) = 0.

Answer

Yes, $ y(1) = 0 $ is satisfied as $ \alpha \to \infty $.

Steps to Solve

Identify the Differential Equation The given initial value problem is ( \frac{dy}{dx} + \alpha y = 0 ). This is a first-order linear differential equation.
Rearrange the Equation We can rewrite the equation in standard form: $$ \frac{dy}{dx} = -\alpha y $$
Separate Variables To solve this, we'll separate the variables ( y ) and ( x ): $$ \frac{dy}{y} = -\alpha dx $$
Integrate Both Sides Integrating both sides gives: $$ \int \frac{1}{y} dy = -\alpha \int dx $$

This simplifies to: $$ \ln |y| = -\alpha x + C $$

Exponentiate to Solve for ( y ) Exponentiating both sides, we have: $$ |y| = e^{- \alpha x + C} = e^C e^{-\alpha x} $$ Setting ( C' = e^C ), we write: $$ y = C' e^{-\alpha x} $$
Use Initial Condition Given the initial condition ( y(0) = 1 ), we substitute ( x = 0 ): $$ 1 = C' e^{0} \implies C' = 1 $$

Thus, the solution simplifies to: $$ y = e^{-\alpha x} $$

Determine ( \alpha ) so that ( y(1) = 0 ) Now we need ( y(1) = 0 ): $$ y(1) = e^{-\alpha \cdot 1} = 0 $$

However, ( e^{-\alpha} ) approaches 0 as ( \alpha \to \infty ). Hence, by choosing ( \alpha = \infty ), we can make ( y(1) = 0 ).

Yes, there is an ( \alpha ) such that ( y(1) = 0 ) as ( \alpha \to \infty ).

More Information

In this case, ( \alpha ) can be made infinitely large to bring the solution ( y ) to zero at ( x = 1 ). The parameter ( \alpha ) controls the rate of decay of ( y ).

Tips

Mistaking ( y(1) = 0 ) as having a finite value of ( \alpha ) instead of realizing it requires ( \alpha ) to approach infinity.

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