Please solve & show steps... **Explanation of Solution Behavior for the Differential Equation**Consider the differential equation:\[ y' = y(8 - ty) \]We are interested in describing how solutions appear to behave as $ t $ increases and how their behavior depends on the initial value $ y_0 $ when $ t = 0 $.### Options:1. **Option A:** For $ y_0 > 0 $, solutions decrease until they intersect the curve $ y = \frac{8}{t} $, then they increase. For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution. For $ y_0 < 0 $, solutions decrease without bound as $ t \to \infty $.2. **Option B:** For $ y_0 > 0 $, solutions increase until they intersect the curve $ y = \frac{8}{t} $, then they decrease. For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution. For $ y_0 < 0 $, solutions increase without bound as $ t \to \infty $.3. **Option C:** For $ y_0 > 0 $, solutions decrease until they intersect the curve $ y = \frac{8}{t} $, then they increase. For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution. For $ y_0 < 0 $, solutions increase without bound as $ t \to \infty $.4. **Option D:** For $ y_0 > 0 $, solutions increase until they intersect the curve $ y = \frac{8}{t} $, then they decrease. For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution. For $ y_0 < 0 $, solutions decrease without bound as $ t \to \infty $.5. **Option E:** For $ y_0 > 0 $, solutions increase without bound as $ t \to \infty $. For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution. For \( y_

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Description: Please solve & show steps... **Explanation of Solution Behavior for the Differential Equation**Consider the differential equation:\[ y' = y(8 - ty) \]We are interested in describing how solutions appear to behave as $ t $ increases and how their be

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Describe how solutions appear to behave as t increases and how their behavior depends on the initial value yo when ! y' = y(8 – ty) 8 For yo > 0 solutions decrease until they intersect the curve y =, then they increase. For yo = 0,y = 0 is an equilibrium solution. For yo < 0, solutions decrease without bound as t → 0. 8 For yo > 0 solutions increase until they intersect the curve y =, then they decrease. For yo = 0,y = 0 is an equilibrium solution. For yo < 0, solutions increase without bound as t → 0. 8 For yo > 0 solutions decrease until they intersect the curve y =, then they increase. For yo = 0,y = 0 is an equilibrium solution. For yo <0, solutions increase without bound as t → 0.

Describe how solutions appear to behave as t increases and how their behavior depends on the initial value yo when ! y' = y(8 – ty) 8 For yo > 0 solutions decrease until they intersect the curve y =, then they increase. For yo = 0,y = 0 is an equilibrium solution. For yo < 0, solutions decrease without bound as t → 0. 8 For yo > 0 solutions increase until they intersect the curve y =, then they decrease. For yo = 0,y = 0 is an equilibrium solution. For yo < 0, solutions increase without bound as t → 0. 8 For yo > 0 solutions decrease until they intersect the curve y =, then they increase. For yo = 0,y = 0 is an equilibrium solution. For yo <0, solutions increase without bound as t → 0.

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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**Explanation of Solution Behavior for the Differential Equation**

Consider the differential equation:

\[ y' = y(8 - ty) \]

We are interested in describing how solutions appear to behave as $ t $ increases and how their behavior depends on the initial value $ y_0 $ when $ t = 0 $.

### Options:
1. **Option A:**

For $ y_0 > 0 $, solutions decrease until they intersect the curve $ y = \frac{8}{t} $, then they increase.

For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution.

For $ y_0 < 0 $, solutions decrease without bound as $ t \to \infty $.

2. **Option B:**

For $ y_0 > 0 $, solutions increase until they intersect the curve $ y = \frac{8}{t} $, then they decrease.

For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution.

For $ y_0 < 0 $, solutions increase without bound as $ t \to \infty $.

3. **Option C:**

For $ y_0 > 0 $, solutions decrease until they intersect the curve $ y = \frac{8}{t} $, then they increase.

For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution.

For $ y_0 < 0 $, solutions increase without bound as $ t \to \infty $.

4. **Option D:**

For $ y_0 > 0 $, solutions increase until they intersect the curve $ y = \frac{8}{t} $, then they decrease.

For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution.

For $ y_0 < 0 $, solutions decrease without bound as $ t \to \infty $.

5. **Option E:**

For $ y_0 > 0 $, solutions increase without bound as $ t \to \infty $.

For $ y_0 = 0 $, $ y = 0 $ is an equilibrium solution.

For \( y_

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