Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

(a) Show that $ϕ (x) = x^{2}$ is an explicit solution to $x \frac{dy}{dx} = 2 y$ on the interval $(- \infty, \infty)$ .
(b) Show that $ϕ (x) = e^{x} - x$ , is an explicit solution to $\frac{dy}{dx} + y^{2} = e^{x} + (1 - 2 x) e^{x} + x^{2} - 1$ on the interval $(- \infty, \infty)$ .
(c) Show that $ϕ (x) = x^{2} - x^{- 1}$ is an explicit solution to $x^{2} \frac{d^{2} y}{{dx}^{2}} = 2 y$ on the interval $(0, \infty)$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1(a): Showing that ϕx=x2 is an explicit solution to xdydx=2y

Firstly, differentiate the given function concerning x.

Therefore, $ϕ' (x) = 2 x,$ for all x on the interval $(- \infty, \infty)$ ,

Substituting the value of for y in the given equation,

$x \frac{dy}{dx} = 2 y x \times ϕ' (x) = 2 \times ϕ (x) x \times (2 x) = 2 \times (x^{2}) 2 x^{2} = 2 x^{2}$

This means, the function $ϕ (x)$ , substituted in the given equation, satisfies the equation for all x in the interval $(- \infty, \infty)$ .

Hence, $ϕ (x) = x^{2}$ is an explicit solution to the equation $x \frac{dy}{dx} = 2 y$ , for all x on the interval $(- \infty, \infty)$ .

Step 2(b): Show that ϕ(x)=ex-x, is an explicit solution to dydx+y2=ex+(1-2x)ex+x2-1a2+b2

Firstly, differentiate the given function concerning x.

Therefore, $ϕ' (x) = e^{x} - 1$ , for all x on the interval $(- \infty, \infty)$ .

Substituting the value of $ϕ (x)$ for y in the L.H.S. (Left-hand side) of the given equation,

$\frac{dy}{dx} + y^{2} = e^{x} - 1 + {(e^{x} - x)}^{2} \frac{dy}{dx} + y^{2} = e^{x} - 1 + e^{2 x} + x^{2} - 2 {xe}^{x} \frac{dy}{dx} + y^{2} = e^{2 x} + e^{x} - 2 {xe}^{x} + x^{2} - 1 \frac{dy}{dx} + y^{2} = e^{2 x} + (1 - 2 x) e^{x} + x^{2} - 1$

Which is the same as the R.H.S. (Right-hand side) of the given equation,

Hence, $ϕ (x) = e^{x} - x$ is an explicit solution to the equation $\frac{dy}{dx} + y^{2} = e^{x} + (1 - 2 x) e^{x} + x^{2} - 1$ , for all x on the interval $(- \infty, \infty)$ .

Step 3(c): Show that ϕx=x2-x-1 is an explicit solution to x2d2ydx2=2y

Firstly, we will differentiate the given function concerning x.

Therefore, $ϕ' (x) = 2 x - (- 1) \times x^{- 2} = 2 x + x^{- 2}$ , for all x in the interval $(0, \infty)$

And $ϕ'' (x) = 2 + (- 2) x^{- 3} = 2 - 2 x^{- 3}$ , for all x in the interval $(0, \infty)$

Substituting the value of for y in the given equation,

$x^{2} \frac{d^{2} y}{{dx}^{2}} = 2 y x^{2} \times ϕ'' (x) = 2 \times ϕ (x) x^{2} \times (2 - 2 x^{- 3}) = 2 \times (x^{2} - x^{- 1}) 2 x^{2} - 2 x^{- 1} = 2 x^{2} - 2 x^{- 1}$

This means, the function $ϕ (x)$ , substituted in the given equation, satisfies the equation, for all x in the interval $(0, \infty)$ .

So, $ϕ (x) = x^{2} - x^{- 1}$ is an explicit solution to the equation $x^{2} \frac{d^{2} y}{{dx}^{2}} = 2 y$ , for all x in the interval $(0, \infty)$ .

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Fundamentals Of Differential Equations And Boundary Value Problems

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Short Answer

Step by Step Solution

Step 1(a): Showing that ϕx=x2 is an explicit solution to xdydx=2y

Step 2(b): Show that ϕ(x)=ex-x, is an explicit solution to dydx+y2=ex+(1-2x)ex+x2-1a2+b2

Step 3(c): Show that ϕx=x2-x-1 is an explicit solution to x2d2ydx2=2y

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

(a) Show that ϕx=x2 is an explicit solution to xdydx=2y on the interval (-∞,∞).(b) Show that ϕ(x)=ex-x, is an explicit solution to dydx+y2=ex+1-2xex+x2-1 on the interval (-∞,∞).(c) Show that ϕx=x2-x-1 is an explicit solution to x2d2ydx2=2y on the interval (0,∞).

Step by Step Solution

Step 1(a): Showing that ϕx=x2 is an explicit solution to xdydx=2y

Step 2(b): Show that ϕ(x)=ex-x, is an explicit solution to dydx+y2=ex+(1-2x)ex+x2-1a2+b2

Step 3(c): Show that ϕx=x2-x-1 is an explicit solution to x2d2ydx2=2y

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