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Fundamentals Of Differential Equations And Boundary Value Problems

R. Kent Nagle, Edward B. Saff, Arthur David Snider

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 616 pages

ISBN: 9780321977069

Chapter 1: Q5 E (page 13)

In Problems 3鈥�8, determine whether the given function is a solution to the given differential equation.
$胃 = 2 e^{3 t} - e^{2 t}$ , $\frac{d^{2} 胃}{{dt}^{2}} - 胃 \frac{诲胃}{dt} + 3 胃 = - 2 e^{2 t}$

Short Answer

Expert verified

The given function is not a solution to the given differential equation.

Step by step solution

01

Differentiating the given equation w.r.t. (with respect to) t

Firstly, we will differentiate $胃 = 2 e^{3 t} - e^{2 t}$ with respect to t,

$\frac{诲胃}{dt} = 6 e^{3 t} - 2 e^{2 t}$

Again, differentiatingwith respect to t,

$\frac{d^{2} 胃}{{dt}^{2}} = 18 e^{3 t} - 4 e^{2 t}$

02

Simplification

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

$\frac{d^{2} 胃}{{dt}^{2}} - 胃 \frac{d 胃}{dt} + 3 胃 = 18 e^{3 t} - 4 e^{2 t} - (2 e^{3 t} - e^{2 t}) 脳 (6 e^{3 t} - 2 e^{2 t}) + 3 脳 (2 e^{3 t} - e^{2 t}) \frac{d^{2} 胃}{{dt}^{2}} - 胃 \frac{d 胃}{dt} + 3 胃 = 18 e^{3 t} - 4 e^{2 t} - (12 e^{6 t} - 4 e^{5 t} - 6 e^{5 t} + 2 e^{4 t}) + 6 e^{3 t} - 3 e^{2 t} \frac{d^{2} 胃}{{dt}^{2}} - 胃 \frac{诲胃}{dt} + 3 胃 = 24 e^{3 t} - 7 e^{2 t} - (12 e^{6 t} - 4 e^{5 t} - 6 e^{5 t} + 2 e^{4 t})$

which is not the same as the R.H.S. (Right-hand side) of the given differential equation.

Hence, $胃 = 2 e^{3 t} - e^{2 t}$ is not a solution to the differential equation $\frac{d^{2} 胃}{{dt}^{2}} - 胃 \frac{诲胃}{dt} + 3 胃 = - 2 e^{2 t}$ .

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Most popular questions from this chapter

Question: Show that, $28 . {鈭�}_{n = 0}^{鈭�} a_{n} x^{n + 1} + {鈭�}_{n = 1}^{鈭�} n b_{n} x^{n - 1} = b_{1} + {鈭�}_{n - 1}^{鈭�} [2 a_{n - 1} + (n + 1) b_{n + 1}] x^{n}$

Decide whether the statement made is True or False. The relation $x^{2} + y^{3} - e^{y} = 1$ is an implicit solution to $\frac{dy}{dx} = \frac{e^{y} - 2 x}{3 y^{2}}$ .

Show that $蠒 (x) = {Ce}^{3 x} + 1$ is a solution tolocalid="1663944867164" $\frac{d y}{d x} - 3 y = - 3$ for any choice of the constant C. Thus, ${Ce}^{3 x} + 1$ is a one-parameter family of solutions to the differential equation. Graph several of the solution curves using the same coordinate axes.

In Problems 21鈥�26, solve the initial value problem.

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