Question

2. (10 points) Consider the initial value problem $\frac{dy}{dt} = \frac{-t + \sqrt{t^2 + 4y}}{2}$, $y(2) = -1$. (a) Show that both $y_1(t) = 1 - t$ and $y_2(t) = -t^2/4$ are solutions to the initial value problem. (This means showing two things: that the function satisfies the DE and satisfies the IC.) (b) Explain why the existence of two solutions to the problem does not contradict the Uniqueness Theorem.

          2. (10 points) Consider the initial value problem
$\frac{dy}{dt} = \frac{-t + \sqrt{t^2 + 4y}}{2}$,
$y(2) = -1$.
(a) Show that both $y_1(t) = 1 - t$ and $y_2(t) = -t^2/4$ are solutions to the initial value problem.
(This means showing two things: that the function satisfies the DE and satisfies the IC.)
(b) Explain why the existence of two solutions to the problem does not contradict the Uniqueness
Theorem.

2. (10 points) Consider the initial value problem
(dy)/(dt) = (-t + √(t^2 + 4y))/(2),
y(2) = -1.
(a) Show that both y1(t) = 1 - t and y2(t) = -t^2/4 are solutions to the initial value problem.
(This means showing two things: that the function satisfies the DE and satisfies the IC.)
(b) Explain why the existence of two solutions to the problem does not contradict the Uniqueness
Theorem.

Added by Colleen L.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Aidan Mcnabb

09/02/2021

Step 1

Step 1: We need to show that both y1(t) = 1 - t and y2(t) = -t^2/4 satisfy the differential equation and the initial condition. Show more…

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2. (10 points) Consider the initial value problem $$\frac{dy}{dt} = \frac{-t + \sqrt{t^2 + 4y}}{2},$$ y(2) = -1. (a) Show that both y1(t) = 1 -t and y2(t) = -t2/4 are solutions to the initial value problem. (This means showing two things: that the function satisfies the DE and satisfies the IC.) (b) Explain why the existence of two solutions to the problem does not contradict the Uniqueness Theorem.