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Chapter 4: Problem 11

Form a differential equation of family of parabolas with focus origin and axis of symmetry along the \(\mathrm{x}\)-axis.

Short Answer

Expert verified
Question: Write the differential equation representing the family of parabolas with their focus at origin and axis of symmetry along the x-axis. Answer: The differential equation for the given family of parabolas is \(2yy' = \frac{y^2}{x}\).

Step by step solution

01

Find the general equation of a parabola with given conditions

Any parabola with focus at (0,0) and axis of symmetry along the x-axis can be represented by the equation \((y-k)^2=4a(x-h)\), where \((h,k)\) is the vertex of the parabola. In this case, we want all parabolas to have their vertex on the x-axis which means the y-coordinate of the vertex will be 0. Therefore, the equation becomes \((y-0)^2=4a(x-h)\) or \(y^2=4a(x-h)\).
02

Write the equation in terms of the arbitrary constant a

We know that the vertex is on the x-axis, so we can write \(y^2=4a(x-h)\) as \(y^2=4a(x-0)\). This allows us to rewrite the equation as \(y^2=4ax\).
03

Differentiate the equation with respect to x

Now, we differentiate the equation \(y^2=4ax\) with respect to x. Using the chain rule, we have \(2yy'=4a\).
04

Eliminate the arbitrary constant a

The arbitrary constant a is present in the equation after differentiation. To eliminate it, we will use the original equation \(y^2=4ax\). We can write \(a=\frac{y^2}{4x}\). Substitute this expression for a in the equation from step 3: \(2yy' = 4\left(\frac{y^2}{4x}\right)\).
05

Simplify and obtain the final differential equation

By simplifying the expression from step 4, we get: \(2yy' = y^2\frac{4}{4x}\) or \(2yy' = y^2\frac{1}{x}\). So, the differential equation representing the family of parabolas with focus at origin and axis of symmetry along the x-axis is: \(2yy' = \frac{y^2}{x}\)

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Most popular questions from this chapter

Prove that the differential equation of the confocal parabolas \(\mathrm{y}^{3}=4 \mathrm{a}(\mathrm{x}+\mathrm{a})\), is \(\mathrm{yp}^{2}+2 \mathrm{xp}-\mathrm{y}=0\), where \(\mathrm{p}=\mathrm{dy} / \mathrm{dx}\) Show that this coincides with the differential equation of the orthogonal curves and interpret the result.

Solve the following differential equations: (i) \(y^{\prime \prime}=x+\sin x\) (ii) \(\mathrm{y}^{\prime \prime}=1+\mathrm{y}^{\prime 2}\) (iii) \(2\left(\mathrm{y}^{\prime}\right)^{2}=\mathrm{y}^{\prime \prime}(\mathrm{y}-1)\) (iv) \(y^{\prime \prime \prime}+y^{\prime \prime 2}=0\)

Verify that \(y=\sin x \cos x-\cos x\) is a solution of the initial value problem \(y^{\prime}+(\tan x) y=\cos ^{2} x\) \(y(0)=-1\), on the interval \(-\pi / 2

(a) Find the general solution \(y_{\mathrm{h}}\) of the homogeneous differential equation \(\frac{d y}{d x}+2 x y=0\) (b) Show that the general solution of the nonhomogeneous equation \(\frac{d y}{d x}+2 x y=3 e^{-x^{2}}\) is equal to the solution \(y_{b}\) in part (a) plus a particular solution to the nonhomogeneous equation.

Solve \(\left(\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{1}{x}+\frac{1}{y}\right) d x\) \(+\left(\frac{y}{\sqrt{x^{2}+y^{2}}}+\frac{1}{y}-\frac{x}{y^{2}}\right) d y=0\)
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