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To solve the separable differential equation we must find two separate integrals: dy= = help (formulas) and d dx help (formulas) The first integral we integrate by substitution: u = help (formulas) du = help (formulas) Solving for y we get one positive solution y = help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(x) = = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs dy 1 y² - 6 dx 2y y(1) = −5.

To solve the separable differential equation we must find two separate integrals: dy= = help (formulas) and d dx help (formulas) The first integral we integrate by substitution: u = help (formulas) du = help (formulas) Solving for y we get one positive solution y = help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(x) = = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs dy 1 y² - 6 dx 2y y(1) = −5.

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.7: Applications
Problem 13EQ
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To solve the separable differential equation we must find two separate integrals: dy= = help (formulas) and d dx help (formulas) The first integral we integrate by substitution: u = help (formulas) du = help (formulas) Solving for y we get one positive solution y = help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(x) = = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs dy 1 y² - 6 dx 2y y(1) = −5.
Transcribed Image Text:To solve the separable differential equation we must find two separate integrals: dy= = help (formulas) and d dx help (formulas) The first integral we integrate by substitution: u = help (formulas) du = help (formulas) Solving for y we get one positive solution y = help (formulas) And one negative solution y = ☐ help (formulas) Note: You must simplify all arbitrary constants down to one constant k. Find the particular solution satisfying the initial condition y(x) = = ☐ help (formulas) Book: Section 1.3 of Notes on Diffy Qs dy 1 y² - 6 dx 2y y(1) = −5.
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