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Chapter 13: Problem 1144

If \((\mathrm{dy} / \mathrm{dx})=\mathrm{y}+3>0\) and \(\mathrm{y}(0)=2\), then \(\mathrm{y}(\log 2)\) is equal to. \(\begin{array}{ll}\text { (A) }-2 & \text { (B) } 5\end{array}\) (C) 7 (D) 13

Short Answer

Expert verified
\(y(\log 2)=\frac{7}{2}\)

Step by step solution

01

Determine the integrating factor

In this step we will find the integrating factor. The given differential equation is a first-order linear ordinary differential equation. The integrating factor is of the form \(e^{\int P(x) dx}\), where \(P(x)\) is the coefficient of the dependent variable in the differential equation. In our case, \(P(x) = 1\). The integrating factor is: \(e^{\int 1 dx} = e^{x}\).
02

Multiply the differential equation by the integrating factor

We will now multiply the given differential equation by the integrating factor to obtain a new differential equation: \((e^x)\frac{\mathrm{dy}}{\mathrm{dx}}+(e^x)y = e^x(\mathrm{y}+3)\) Notice that the left-hand side of the new differential equation is the derivative of the product of \(y(x)\) and the integrating factor. \(\frac{d}{dx}(ye^x) = e^x(y+3)\)
03

Integrate both sides of the equation

Now we will integrate both sides of the equation: \(∫\frac{d}{dx}(ye^x)dx = ∫e^x(y+3)dx\) After integrating, we get: \(ye^x = e^x(y + 3) + C\), where C is the integration constant. Now we'll solve for y: \(y(x) = \frac{e^x(y+3) + C}{e^x}\)
04

Apply the initial condition

Now we will use the initial condition \(y(0) = 2\) to find the value of C: \(2 = \frac{e^0(2+3) + C}{e^0}\) Simplifying, we get: \(2 = 5 + C\) \(C = -3\) Therefore, \(y(x) = \frac{e^x(y+3) - 3}{e^x}\)
05

Evaluate y at x = \(\log 2\)

Finally, we will find the value of \(y(\log 2)\) using the function we found: \(y(\log 2) = \frac{e^{\log 2}(y+3) - 3}{e^{\log 2}}\) Recall that \(e^{\log 2} = 2\), \(y(\log 2) = \frac{2(y+3) - 3}{2}\) \(y(\log 2) = \frac{2(5) - 3}{2}\) \(y(\log 2) = \frac{7}{2}\) None of the given options match the derived answer, \(\frac{7}{2}\). There may be an error in the given options.

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Most popular questions from this chapter

Integrating factor of differential equation \([1 /(\cos x)] \cdot(d y / d x)+[1 /(\sin x)] y=1\) is. (A) \(\sec x\) (B) \(\cos \mathrm{x}\) (C) \(\tan \mathrm{x}\) (D) \(\sin \mathrm{x}\)

Family of curves \(\mathrm{y}=\mathrm{e}^{\mathrm{x}}(\mathrm{A} \cos \mathrm{x}+\mathrm{B} \sin \mathrm{x})\) represents the differential equation: \(\quad\) (where \(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constant) (A) \(\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}\right)-2(\mathrm{dy} / \mathrm{dx})+\mathrm{y}=0\) (B) \(\left(d^{2} y / d x^{2}\right)-2(d y / d x)-2 y=0\) (C) \(\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}\right)-2(\mathrm{dy} / \mathrm{dx})-\mathrm{y}=0\) (D) \(\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}\right)-2(\mathrm{dy} / \mathrm{dx})+2 \mathrm{y}=0\)

Solution of \((\mathrm{y} / \mathrm{x}) \cos (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})-(\mathrm{y} / \mathrm{x})]\) \(+\sin (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})+(\mathrm{y} / \mathrm{x})]=0 ; \mathrm{y}(1)=(\pi / 2)\) is: (A) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 2 \mathrm{x})\) (B) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / \mathrm{x})\) (C) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 3 \mathrm{x})\) (D) none of these

Let \(\mathrm{m}\) and \(\mathrm{n}\) be respectively the degree and order of the differential equation of whose solution is \(\mathrm{y}=\mathrm{cx}+\mathrm{c}^{2}-3 \mathrm{c}^{(3 / 2)}+2\) where \(\mathrm{c}\) is parameter is (A) \(\mathrm{m}=1, \mathrm{n}=4\) (B) \(\mathrm{m}=1, \mathrm{n}=4\) (C) \(\mathrm{m}=2, \mathrm{n}=2\) (D) \(\mathrm{m}=4, \mathrm{n}=1\)

If \(\mathrm{x}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}(\log \mathrm{y}-\log \mathrm{x}+1)\), then the solution of the equation is: (A) \(x \log (\mathrm{y} / \mathrm{x})=\mathrm{cy}\) (B) \(\log (\mathrm{y} / \mathrm{x})=\mathrm{cx}\) (C) \(\log (\mathrm{x} / \mathrm{y})=\mathrm{cy}\) (D) \(\mathrm{y} \cdot \log (\mathrm{x} / \mathrm{y})=\mathrm{cx}\)
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