/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Solve the following initial valu... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the following initial value problems. $$y^{\prime}(t)-2 y=8, y(0)=0$$

Short Answer

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Question: Determine the particular solution for the given initial value problem involving a first-order linear differential equation: $y'(t) - 2y(t) = 8$, with an initial condition $y(0) = 0$. Answer: The particular solution for the given initial value problem is $y(t) = -4 + 4e^{2t}$.

Step by step solution

01

Identify the type of differential equation

This is a first-order linear differential equation in the form: $$y'(t) + P(t)y(t) = Q(t)$$ Where \(P(t) = -2\) and \(Q(t) = 8\).
02

Find the integrating factor

We need to find the integrating factor, which is given by: $$\mu(t) = e^{\int P(t) dt}$$ In our case, \(P(t) = -2\), so we have: $$\mu(t) = e^{\int -2 dt} = e^{-2t}$$
03

Multiply both sides by the integrating factor

Multiply both sides of the equation by the integrating factor \(\mu(t) = e^{-2t}\): $$(e^{-2t}y'(t)) - 2e^{-2t}y(t) = 8e^{-2t}$$ The left side of this equation is now the derivative of the product \(\mu(t)y(t)\): $$\frac{d}{dt}(e^{-2t}y(t)) = 8e^{-2t}$$
04

Integrate both sides with respect to t

Now, we integrate both sides with respect to \(t\): $$\int\frac{d}{dt}(e^{-2t}y(t)) dt = \int 8e^{-2t} dt$$ This yields: $$e^{-2t}y(t) = -4e^{-2t} + C$$
05

Solve for y(t)

Now, we solve for \(y(t)\) by multiplying both sides by \(e^{2t}\): $$y(t) = -4 + Ce^{2t}$$
06

Apply the initial condition to find the particular solution

Finally, to find the particular solution, substitute the initial condition \(y(0) = 0\) into the form of the general solution: $$0 = -4 + Ce^{2 \times 0} \Rightarrow 0 = -4 + C$$ This gives \(C = 4\), so the particular solution is: $$y(t) = -4 + 4e^{2t}$$

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Most popular questions from this chapter

One possible model that describes the free fall of an object in a gravitational field subject to air resistance uses the equation \(v^{\prime}(t)=g-b v,\) where \(v(t)\) is the velocity of the object for \(t \geq 0\), \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, and \(b>0\) is a constant that involves the mass of the object and the air resistance. a. Verify by substitution that a solution of the equation, subject to the initial condition \(v(0)=0,\) is \(v(t)=\frac{g}{b}\left(1-e^{-b t}\right)\). b. Graph the solution with \(b=0.1 s^{-1}\). c. Using the graph in part (c), estimate the terminal velocity \(\lim _{t \rightarrow \infty} v(t)\).

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