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Chapter 6: Problem 5

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
All given solutions are correct solutions for the given differential equations when substituted and checked within their respective equations.

Step by step solution

01

Solution 1 Verification

Differentiate \(y=C e^{4x}\) using the chain rule to get the derivative \(y'=4Ce^{4x}\). Substitute \(y'\) and \(y\) into the differential equation \(y'=4y\). If \(4Ce^{4x} = 4*C e^{4x}\) holds true, then \(y=C e^{4x}\) is a solution.
02

Solution 2 Verification

Differentiate \(y=e^{-x}\) using the chain rule to get \(y'=-e^{-x}\). Substitute into the differential equation \(3y'+4y=e^{-x}\). If \(-3e^{-x} + 4e^{-x} = e^{-x}\) holds, then \(y=e^{-x}\) is indeed a solution.
03

Solution 3 Verification

Rearrange the equation \(x^{2}+y^{2}=Cy\) into the differential form \(y'=2xy/(x^{2}-y^{2})\). Substitute \(y\) and \(y'\) into the differential equation. If it produces a true statement, then the given function is the solution.
04

Solution 4 Verification

Rewrite \(y^{2}-2 \ln y=x^{2}\) as \( \frac{d y}{d x}=\frac{x y}{y^{2}-1}\) . Substitute \(y\) and \(y'\) into the differential equation and see if it holds true.
05

Solution 5 Verification

Differentiate \(y=C_{1} \cos x+C_{2} \sin x\) twice to get second derivative. Substitute into \(y^{\prime \prime}+y=0\). If equation holds, then the function is correct solution.
06

Solution 6 Verification

Differentiate \(y=C_{1} e^{-x} \cos x+C_{2} e^{-x} \sin x\) twice to get second derivative. Substitute into the differential equation \(y^{\prime \prime}+2 y^{\prime}+2 y=0\). If equation holds true, then function is correct solution.
07

Solution 7 Verification

To verify \(y=-\cos x \ln |\sec x+\tan x|\), differentiate it twice to get the second derivative. Substitute the function and its second derivative in the equation \(y^{\prime \prime}+y=\tan x\). If it holds true, then the solution is correct.
08

Solution 8 Verification

To verify \(y=\frac{2}{3}(e^{-2x}+e^{x})\), differentiate it twice to get the second derivative. Substitute the function and its second derivative into the equation \(y^{\prime \prime}+2 y^{\prime}=2e^{x}\). If it holds true, then the solution is correct.

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Most popular questions from this chapter

The logistic equation models the growth of a population. Use the equation to (a) find the value of \(k\), (b) find the carrying capacity, (c) find the initial population, (d) determine when the population will reach \(\mathbf{5 0 \%}\) of its carrying capacity, and (e) write a logistic differential equation that has the solution \(P(t)\). \(P(t)=\frac{1500}{1+24 e^{-0.75 t}}\)

Find the logistic equation that satisfies the initial condition. \(\frac{d y}{d t}=\frac{4 y}{5}-\frac{y^{2}}{150}\) \((0,8)\)

(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. $$ \begin{array}{lll} \text { Differential Equation } & \text { Initial Condition } & x \text { -value } \\ \frac{d y}{d x}+6 x y^{2}=0 & (0,3) & x=1 \end{array} $$

Find the logistic equation that satisfies the initial condition. \(\frac{d y}{d t}=y\left(1-\frac{y}{40}\right) \quad(0,8)\)

The table shows the net receipts and the amounts required to service the national debt (interest on Treasury debt securities) of the United States from 1992 through 2001 . The monetary amounts are given in billions of dollars. (Source: U.S. Office of Management and Budget) $$ \begin{aligned} &\begin{array}{|l|c|c|c|c|c|} \hline \text { Year } & 1992 & 1993 & 1994 & 1995 & 1996 \\ \hline \text { Receipts } & 1091.3 & 1154.4 & 1258.6 & 1351.8 & 1453.1 \\ \hline \text { Interest } & 292.3 & 292.5 & 296.3 & 332.4 & 343.9 \\ \hline \end{array}\\\ &\begin{array}{|l|c|c|c|c|c|} \hline \text { Year } & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \text { Receipts } & 1579.3 & 1721.8 & 1827.5 & 2025.2 & 1991.2 \\ \hline \text { Interest } & 355.8 & 363.8 & 353.5 & 361.9 & 359.5 \\ \hline \end{array} \end{aligned} $$ (a) Use the regression capabilities of a graphing utility to find an exponential model \(R\) for the receipts and a quartic model \(I\) for the amount required to service the debt. Let \(t\) represent the time in years, with \(t=2\) corresponding to 1992 . (b) Use a graphing utility to plot the points corresponding to the receipts, and graph the corresponding model. Based on the model, what is the continuous rate of growth of the receipts? (c) Use a graphing utility to plot the points corresponding to the amount required to service the debt, and graph the quartic model. (d) Find a function \(P(t)\) that approximates the percent of the receipts that is required to service the national debt. Use a graphing utility to graph this function.
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