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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: Q7RP (page 1)

Decide whether the statement made is True or False. The function $x (t) = t^{- 3} \sin t + 4 t^{- 3}$ is a solution to $t^{3} \frac{dx}{dt} = \cos t - 3 t^{2} x$ .

Short Answer

Expert verified

The statement is true.

Step by step solution

01

Finding the derivative of the equation x(t)=t-3sin t+4t-3 with respect to t.

Consider, $x (t) = t^{- 3} \sin t + 4 t^{- 3}$

Then,

$\frac{dx}{dt} = \frac{d}{dt} (t^{- 3} sint + 4 t^{- 3}) = t^{- 3} . \frac{d}{dt} (sint) + sint \frac{d}{dt} (t^{- 3}) + \frac{d}{dt} (4 t^{- 3}) = t^{- 3} 路 \cos t + \sin t (- 3 t^{- 4}) - 12 t^{- 4}$

02

Putting the value of dxdt in the LHS of the equation t3dxdt=cos t-3t2x to check if RHS is obtained

Equating Left and Right-hand side,

$LHS = t^{3} \frac{dx}{dt} = t^{3} (t^{- 3} 路 \cos t + \sin t (- 3 t^{- 4}) - 12 t^{- 4}) = \cos t - 3 t^{- 1} 路 \sin t - 12 t^{- 1}$

$RHS = \cos t - 3 t^{2} x = cost - 3 t^{2} (t^{- 3} \sin t + 4 t^{- 3}) = cost - 3 t^{- 1} 路 \sin t - 12 t^{- 1}$

Hence, LHS = RHS

Therefore, this shows that $x (t) = t^{- 3} sint + 4 t^{- 3}$ is a solution to $t^{3} \frac{dx}{dt} = cost - 3 t^{2} x$ .

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Most popular questions from this chapter

In Problems 13-16, write a differential equation that fits the physical description. The rate of change of the mass A of salt at time t is proportional to the square of the mass of salt present at time t.

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$x = 2 \cos t - 3 \sin t 鈥�, 鈥� 鈥� 鈥� x'' + x = 0$

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$x \frac{d^{2} y}{d x^{2}} + 3 x - \frac{d y}{d x} = e^{3 x}$

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