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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

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

If \(y=\left[x+\sqrt{ \left.\left(1+x^{2}\right)\right]^{n}}\right.\), then \(\left(1+x^{2}\right) \cdot\left(d^{2} y / d x^{2}\right)+x \cdot(d y / d x)\) - (A) - (B) \(2 \mathrm{x}^{2} \mathrm{y}\) (C) \(\mathrm{n}^{2} \mathrm{y}\) (D) \(-\mathrm{n}^{2} \mathrm{y}\)

The differential equation whose solution is \(\mathrm{Ax}^{2}+\mathrm{By}^{2}=1\), where \(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constants is of. (A) second order and second degree (B) first order and first degree (C) first order and second degree (D) second order and first degree

The solution of \(\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}\right)=0\) represents: (A) a point (B) a straight line (C) a parabola (D) a circle

The differential equation of all circles passing through the origin and having their centers on the y-axis is: OR The differential equation for the family of curves \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ay}=0\), where a is an arbitrary constant is: (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (C) \(2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (D) \(\left(x^{2}+y^{2}\right) y^{1}=2 x y\)
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