Consider the Bernoulli I.V.P zy' – x?yn = y, y(1) = 1. where n=3.%3D%3DLet v = y™ be the substitution that converts the above equation into linear, find m.find v(2). [ Enter fraction of the form a/b or number up to 3 decimal digits]

Answered: Consider the Bernoulli I.V.P ry' – x?y"…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.7: More On Inequalities

Problem 44E

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Question

Consider the Bernoulli I.V.P zy' – x?yn = y, y(1) = 1. where n=3.
%3D
%3D
Let v = y™ be the substitution that converts the above equation into linear, find m.
find v(2). [ Enter fraction of the form a/b or number up to 3 decimal digits]

Expert Solution

Step 1

Given

$x y' - x^{2} y^{n} = y$

y(1) = 1

where n = 3

Step 2

Substituting the value of n = 3 in the $x y' - x^{2} y^{n} = y$

$x y' - x^{2} y^{3} = y$

$x y' - y = x^{2} y^{3}$

Dividing both sides by x and y³

$y^{- 3} y' - \frac{y^{- 2}}{x} = x$

Let's assume v = y^-2

Differentiating both sides with respect to x

$\frac{d v}{d x} = - 2 y^{- 2 - 1} y'$ (because d(xⁿ)/dx = nx^n-1 and d(constant)/dx = 0)

$- \frac{1}{2} \frac{d v}{d x} = y^{- 3} y'$

Substituting $- \frac{1}{2} \frac{d v}{d x} = y^{- 3} y'$ and v = y^-2 in the $y^{- 3} y' - \frac{y^{- 2}}{x} = x$

$- \frac{1}{2} \frac{d v}{d x} - \frac{v}{x} = x$

$\frac{d v}{d x} + \frac{2 v}{x} = - 2 x$

Now it is a linear differential equation

So, comparing v = y^-2 with v = y^m

then

m = -2 Answer

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Consider the Bernoulli I.V.P ry' – x?y" – y, y(1) = 1. where n=3. Let v = y™ be the substitution that converts the above equation into linear, find m. find v(2). [ Enter fraction of the form a/b or number up to 3 decimal digits]