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Chapter 5: Problem 58

\(y^{2}-2 y-8 x+17=0\) \(2 y \frac{d y}{d x}-2 \frac{d y}{d x}-8=0\) \(\frac{d y}{d x}=\frac{y}{y-1}=1\) \(\Rightarrow y=5, \quad x=4\)

Short Answer

Expert verified
#Answer# x = 4, y = 5

Step by step solution

01

Solve the differential equation

We are given the following first-order differential equation: $$2 y \frac{d y}{d x}-2 \frac{d y}{d x}-8=0$$ Group the terms containing dy/dx and add 8 to both sides: $$(2 y - 2) \frac{d y}{d x} = 8$$ Now, separate the variables and integrate both sides: $$\frac{d y}{2 y - 2} = \frac{8}{2} dx$$ Integrating: $$\int \frac{d y}{2 y - 2} = 4 \int dx$$
02

Find the implicit solution for y

To find the implicit solution for y, integrate both sides: $$\frac{1}{2} \int \frac{2 dy}{2 y - 2} = \int 4 dx$$ $$\frac{1}{2} \ln|2 y - 2| = 4 x + C$$ To solve for y, first multiply both sides by 2: $$\ln|2 y - 2| = 8 x + 2C$$ Now, take the exponent of both sides to remove the natural log: $$2 y - 2 = e^{8 x + 2C}$$ Add 2 to both sides: $$2 y = e^{8 x + 2C} + 2$$ Finally, divide by 2: $$y = \frac{e^{8 x + 2C} + 2}{2}$$
03

Substitute the provided particular solution into the implicit solution

We are given the following particular solution: $$\frac{d y}{d x}=\frac{y}{y-1}=1$$ Now substitute y = 5 and x = 4: $$1 = \frac{5}{5-1}$$ Since the given particular solution is true, we can confirm that y = 5 and x = 4.

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Most popular questions from this chapter

\(y=\ln x\) \(\frac{d y}{d x}=\frac{1}{x}\) Slope of normal \(\Rightarrow-x_{1}\) Slope of 1 chord \(=\frac{1}{e-1}\) \(\Rightarrow \frac{x_{1}}{e-1}=1\) \(\Rightarrow \quad x_{1}=e-1\) \(y_{1}=\ln (e-1)\)

\(f_{1}^{\prime}(x)=2 x-1 \quad \& \quad f_{2}^{\prime}(x)=3 x^{2}-2 x-2\) \(\Rightarrow 2 x_{1}-1=3 x_{2}^{2}-2 x_{2}-2\) \(\Rightarrow 3 x_{2}^{2}-2 x_{2}-2 x_{1}-1=0\) For \(\mathrm{x}_{2}\) be real, \(\mathrm{D} \geq 0\) \(4-4\left(2 x_{1}+1\right)(3)\) \(4-24 x_{1}-12\) \(\Rightarrow-\left(24 x_{1}+8\right)\) There can be infinite such values of \(\mathrm{x}_{\mathrm{l}}\)

\(y=\left(\frac{x}{2}-a\right)^{2}+a-2\) \(4(y-(a-2))=(x-2 a)^{2}\) Vertex \(\Rightarrow \mathrm{h}=2 \mathrm{a}, \mathrm{k}=\mathrm{a}-2\) Locus of vertex \(\Rightarrow \mathrm{y}=\frac{\mathrm{x}}{2}-2\) \(\Rightarrow 2 y=x-4\)

Slope of secant \(=\frac{9 a^{2}-a^{2}}{3 a-a}=4 a\) \(\frac{d y}{d x}=2 x=4 a\) \(x=2 a\) \(y=4 a^{2}\)

eqn of normal \(y=m x-2 m-m^{3}\) It passes through \((6,0)\) \(\mathrm{m}^{3}-4 \mathrm{~m}=0\) \(\mathrm{m}\left(\mathrm{m}^{2}-4\right)=0\) \(\mathrm{m}=0, \pm 2\) Pts are \(\left(a m^{2}-2 a m\right)\) \((0,0),(4,-4),(4,4)\) Minimum distance \(=\sqrt{21}-\sqrt{5}\)
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