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Chapter 8: Q77SE (page 359)

In Problems 25鈥�28 use (12) to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

\(x\frac{{dy}}{{dx}} - 3xy = 1;y = {e^{3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} dt\)

Short Answer

Expert verified

The indicated function is a solution of the differential function.

Step by step solution

01

Simplify the given differential equation.

Let the given differential equation be \(y = {e^{3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} dt\).

Multiply each side of the equation by \({e^{ - 3x}}\).

\(\begin{array}{l}y{e^{ - 3x}} = {e^{3x}}{e^{ - 3x}}\int_1^x {\frac{{{e^{ - 3t}}}}{t}} \;dt\\y{e^{ - 3x}} = \int_1^x {\frac{{{e^{ - 3t}}}}{t}} \;dt\end{array}\)

02

Determine the solution of the indicated function.

Take differential on both sides of the equation.

Multiply \(x\) on both sides of the equation.

\(x{e^{ - 3x}}\frac{{dy}}{{\;dx}} - 3yx{e^{ - 3x}} = {e^{ - 3x}}\)

Divide\({e^{ - 3x}}\)on both sides of the equation.

\(x\frac{{dy}}{{\;dx}} - 3yx = 1\)

Hence, the indicated function is a solution of the differential function.

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Most popular questions from this chapter

The following observations are on stopping distance (ft) of a particular truck at \(20mph\) under specified experimental conditions (鈥淓xperimental Measurement of the Stopping Performance of a Tractor-Semitrailer from Multiple Speeds,鈥� NHTSA, DOT HS 811 488, June 2011):

\(32.1 30.6 31.4 30.4 31.0 31.9\)

The cited report states that under these conditions, the maximum allowable stopping distance is \(30\). A normal probability plot validates the assumption that stopping distance is normally distributed.

a. Does the data suggest that true average stopping distance exceeds this maximum value? Test the appropriate hypotheses using \(\alpha = .01\).

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c. Repeat (b) using \(\sigma = .80\) and compare to the results of (b).

d. What sample size would be necessary to have \(\alpha = .01\) and \(\beta = .10\) when \(\mu = 31\) and \(\sigma = .65\)?

The accompanying data on cube compressive strength (MPa) of concrete specimens appeared in the article 鈥淓xperimental Study of Recycled Rubber-Filled High-Strength Concrete鈥� (Magazine of Concrete Res., 2009: 549鈥�556):

\(\begin{array}{l}112.3 97.0 92.7 86.0 102.0\\99.2 95.8 103.5 89.0 86.7\end{array}\)

a. Is it plausible that the compressive strength for this type of concrete is normally distributed?

b. Suppose the concrete will be used for a particular application unless there is strong evidence that true average strength is less than \(100MPa\). Should the concrete be used? Carry out a test of appropriate hypotheses.

Reconsider the accompanying sample data on expense ratio (%) for large-cap growth mutual funds first introduced in Exercise 1.53.

\(\begin{array}{l}0.52 1.06 1.26 2.17 1.55 0.99 1.10 1.07 1.81 2.05\\0.91 0.79 1.39 0.62 1.52 1.02 1.10 1.78 1.01 1.15\end{array}\)

A normal probability plot shows a reasonably linear pattern.

a. Is there compelling evidence for concluding that the population mean expense ratio exceeds \(1\% \)? Carry out a test of the relevant hypotheses using a significance level of \(.01\).

b. Referring back to (a), describe in context type I and II errors and say which error you might have made in reaching your conclusion. The source from which the data was obtained reported that \(\mu = 1.33\) for the population of all \(762\) such funds. So, did you actually commit an error in reaching your conclusion?

c. Supposing that \(\sigma = .5\), determine and interpret the power of the test in (a) for the actual value of m stated in (b).

The article 鈥淯ncertainty Estimation in Railway Track Life-Cycle Cost鈥� (J. of Rail and Rapid Transit, 2009) presented the following data on time to repair (min) a rail break in the high rail on a curved track of a certain railway line.

\(159 120 480 149 270 547 340 43 228 202 240 218\)

A normal probability plot of the data shows a reasonably linear pattern, so it is plausible that the population distribution of repair time is at least approximately normal. The sample mean and standard deviation are \(249.7\) and \(145.1\), respectively.

a. Is there compelling evidence for concluding that true average repair time exceeds \(200\) min? Carry out a test of hypotheses using a significance level of \(.05\).

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