Use the substitutionu(x)%3D2х -Зу(x)+2to solve the following IVP:У'+ (2х - Зу +2)23D0, у (0) %— 2

Given that diffrancial equation y'+2x-3y+22=0 , y0=2Solutiondydx=-2x-3y+22let…

Answered: Use the substitution ux)%3D2х - Зу…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Question

Use the substitution
u(x)%3D2х -Зу(x)+2
to solve the following IVP:
У'+ (2х - Зу +2)23D0, у (0) %— 2

Expert Solution

Step 1

$G i v e n t h a t d i f f r a n c i a l e q u a t i o n y' + {(2 x - 3 y + 2)}^{2} = 0, y (0) = 2 S o l u t i o n \frac{d y}{d x} = - {(2 x - 3 y + 2)}^{2} l e t u = 2 x - 3 y + 2 \frac{d u}{d x} = 2 - 3 \frac{d y}{d x} \frac{d y}{d x} = \frac{2}{3} - \frac{d u}{3 d x} \frac{2}{3} - \frac{d u}{3 d x} = - u^{2} \frac{d u}{d x} = 2 + 3 u^{2} \int \frac{d u}{2 + 3 u^{2}} = \int d x \int \frac{d u}{{\sqrt{2}}^{2} + {(\sqrt{3} u)}^{2}} = \int d x \frac{1}{\sqrt{2} \times \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} u}{\sqrt{2}}) = x + c y (0) = 2 \frac{1}{\sqrt{2} \times \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (2 x - 3 y + 2)}{\sqrt{2}}) = x + c \frac{1}{\sqrt{2} \times \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (2 (0) - 3 (2) + 2)}{\sqrt{2}}) = 0 + c \frac{1}{\sqrt{6}} \tan^{- 1} (\frac{\sqrt{3} (- 4)}{\sqrt{2}}) = c c = - \frac{1}{\sqrt{6}} \tan^{- 1} (2 \sqrt{6}) \frac{1}{\sqrt{2} \times \sqrt{3}} \tan^{- 1} (\frac{\sqrt{3} (2 x - 3 y + 2)}{\sqrt{2}}) = x - \frac{1}{\sqrt{6}} \tan^{- 1} (2 \sqrt{6})$