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Fundamentals Of Differential Equations And Boundary Value Problems

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

$Math Studyset Vaia Explanations$ Math

9th Edition · 616 pages

ISBN: 9780321977069

Chapter 1: Q4 E (page 13)

In Problems 3–8, determine whether the given function is a solution to the given differential equation.
$x = 2 \cos t - 3 \sin t, x'' + x = 0$

Short Answer

Expert verified

The given function is a solution to the given differential equation.

Step by step solution

01

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

Firstly, we will differentiate $x = 2 \cos t - 3 \sin t$ with respect to t,

$x' = - 2 \sin t - 3 \cos t$

Again, differentiating with respect to t,

$x'' = - 2 \cos t - 3 (- \sin t) x'' = - 2 \cos t + 3 \sin t$

02

Simplification

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

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

which is the same as the R.HS. (Right-hand side) of the given differential equation.

Hence, $x = 2 \cos t - 3 \sin t$ is a solution to the differential equation $x'' + x = 0$ .

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Most popular questions from this chapter

Variation of Parameters. Here is another procedure for solving linear equations that is particularly useful for higher-order linear equations. This method is called variation of parameters. It is based on the idea that just by knowing the form of the solution, we can substitute into the given equation and solve for any unknowns. Here we illustrate the method for first-order equations (see Sections 4.6 and 6.4 for the generalization to higher-order equations).

(a) Show that the general solution to (20) $\frac{dy}{dx} + P (x) y = Q (x)$ has the form $y (x) = C y_{h} (x) + y_{p} (x)$ ,where $y_{h}$ ( $\neq 0$ is a solution to equation (20) when $Q (x) = 0$ ,

C is a constant, and $y_{p} (x) = v (x) y_{h} (x)$ for a suitable function v(x). [Hint: Show that we can take $y_{h} = μ^{- 1} (x)$ and then use equation (8).] We can in fact determine the unknown function $y_{h}$ by solving a separable equation. Then direct substitution of v $y_{h}$ in the original equation will give a simple equation that can be solved for v.

Use this procedure to find the general solution to (21) localid="1663920708127" $\frac{dy}{dx} + \frac{3}{x} y = x^{2}$ , x > 0 by completing the following steps:

(b) Find a nontrivial solution $y_{h}$ to the separable equation (22) localid="1663920724944" $\frac{dy}{dx} + \frac{3}{x} y = 0$ , localid="1663920736626" $x > 0$ .

(c) Assuming (21) has a solution of the formlocalid="1663920777078" $y_{p} (x) = v (x) y_{h} (x)$ , substitute this into equation (21), and simplify to obtain localid="1663920789271" $v' (x) = \frac{x^{2}}{y_{h} (x)}$ .

d) Now integrate to getlocalid="1663920800433" $v (x)$

(e) Verify thatlocalid="1663920811828" $y (x) = C y_{h} (x) + v (x) y_{h} (x)$ is a general solution to (21).

In problems $1 - 4$ Use Euler’s method to approximate the solution to the given initial value problem at the points x = 0.1, 0.2, 0.3, 0.4, and 0.5, using steps of size 0.1 (h = 0.1).

$\frac{dy}{dx} = x + y, y (0) = 1$

In problems 1-6, identify the independent variable, dependent variable, and determine whether the equation is linear or nonlinear.

$5 \frac{dx}{dt} + 5 x^{2} + 3 = 0$

Verify that the function $ϕ (x) = c_{1} e^{x} + c_{2} e^{- 2 x}$ is a solution to the linear equation $\frac{d^{2} y}{{dx}^{2}} + \frac{dy}{dx} - 2 y = 0$ for any choice of the constants $c_{1}$ and $c_{2}$ . Determine $c_{1}$ and $c_{2}$ so that each of the following initial conditions is satisfied.

(a) $y (0) = 2, y' (0) = 1$

(b) $y (1) = 1, y' (1) = 0$

Question: The Taylor series for f(x) =ln (x)about x₂=0given in equation (13) can also be obtained as follows:

(a)Starting with the expansion 1/ (1-s) = $\sum_{n = 0}^{\infty} s''$ and observing that

'

obtain the Taylor series for 1/xabout x₀= 1.

(b)Since use the result of part (a) and termwise integration to obtain the Taylor series for f (x)=lnxaboutx₀= 1.

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