/*! 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 3 Verify the solution of the diffe... [FREE SOLUTION] | 91影视

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Chapter 6: Problem 3

Verify the solution of the differential equation. Solution 1\. \(y=C e^{4 x}\) 2\. \(y=e^{-x}\) 3\. \(x^{2}+y^{2}=C y\) 4\. \(y^{2}-2 \ln y=x^{2}\) 5\. \(y=C_{1} \cos x+C_{2} \sin x\) 6\. \(y=C_{1} e^{-x} \cos x+C_{2} e^{-x} \sin x\) 7\. \(y=-\cos x \ln |\sec x+\tan x|\) 8\. \(y=\frac{2}{3}\left(e^{-2 x}+e^{x}\right)\) Differential Equation \(y^{\prime}=4 y\) \(3 y^{\prime}+4 y=e^{-x}\) \(y^{\prime}=2 x y /\left(x^{2}-y^{2}\right)\) \(\frac{d y}{d x}=\frac{x y}{y^{2}-1}\) \(y^{\prime \prime}+y=0\) \(y^{\prime \prime}+2 y^{\prime}+2 y=0\) \(y^{\prime \prime}+y=\tan x\) \(y^{\prime \prime}+2 y^{\prime}=2 e^{x}\)

Short Answer

Expert verified
Solutions 1, 2, 5, and 6 are correct

Step by step solution

01

Solution 1

Start with the differential equation \(y' = 4y\). If we differentiate the proposed solution \(y = Ce^{4x}\), we get \(y' = 4Ce^{4x}\), which can be simplified as \(y' = 4y\). This solution is verified.
02

Solution 2

Start with the differential equation \(3y' + 4y = e^{-x}\). If we differentiate the proposed solution \(y = e^{-x}\), we get \(y' = -e^{-x}\). Plugging \(y\) and \(y'\) into the differential equation, we get \(3(-e^{-x}) + 4e^{-x} = e^{-x}\), which simplifies to \(e^{-x} = e^{-x}\). This solution is verified.
03

Solution 3

Start with the differential equation \(y' = 2xy / (x^2 - y^2)\). If we differentiate the proposed solution \(x^2 + y^2 = Cy\), we get \(y' = 2x - 2Cy / (x^2 - y^2)\). This does not satisfy the differential equation, hence this solution is not correct.
04

Solution 4

Start with the differential equation \(dy/dx = xy / (y^2 - 1)\). If we differentiate the proposed solution \(y^2 - 2lny = x^2\), we get \(dy/dx = 2x - 2/y/(y^2 - 1)\). This does not satisfy the differential equation, hence this solution is not correct.
05

Solution 5

Analyzing fifth equation, starting from \(y''+y=0\); differentiate the proposed solution \(y = C1cosx + C2sinx\), to get \(y'' = - C1cosx - C2sinx\), it fulfills aforementioned equation thus the solution is correct.
06

Solution 6

For the equation \(y'' + 2y' + 2y = 0\), differentiate \(y = C1e^{-x}cosx + C2e^{-x}sinx\) to get \(y'' + 2y' + 2y = 0\), making aforementioned differential equation correct.
07

Solution 7

Consider equation \(y'' + y = tanx\), if we differentiate \(y =- cosx ln|secx + tanx|\), we don鈥檛 get \(y'' + y = tanx\). Hence, this solution is not verified.
08

Solution 8

For differential equation \(y'' + 2y' = 2e^x\), when deriving \(y = 2/3*(e^{-2x} + e^x)\), it doesn't fulfill the original equation. Therefore, this solution is not correct.

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Most popular questions from this chapter

Solve the Bernoulli differential equation. $$ y^{\prime}+x y=x y^{-1} $$

It is known that \(y=e^{k t}\) is a solution of the differential equation \(y^{\prime \prime}-16 y=0 .\) Find the values of \(k\).

(a) use Euler's Method with a step size of \(h=0.1\) to approximate the particular solution of the initial value problem at the given \(x\) -value, (b) find the exact solution of the differential equation analytically, and (c) compare the solutions at the given \(x\) -value. $$ \begin{array}{lll} \text { Differential Equation } & \text { Initial Condition } & x \text { -value } \\ \frac{d y}{d x}=2 x\left(1+y^{2}\right) & (1,0) & x=1.5 \end{array} $$

Find the orthogonal trajectories of the family. Use a graphing utility to graph several members of each family. \(x^{2}-2 y^{2}=C\)

Let \(f\) be a twice-differentiable real-valued function satisfying \(f(x)+f^{\prime \prime}(x)=-x g(x) f^{\prime}(x)\) where \(g(x) \geq 0\) for all real \(x\). Prove that \(|f(x)|\) is bounded.
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