The family of curves represented respectively by the solutions of the differential equations $\frac{d y}{d x} = \frac{x^{2} + x + 1}{y^{2} + y + 1}$ and $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$

Touch each other

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Are orthogonal

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Are identical

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Neither touch nor intersect

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Solution

The correct option is B Are orthogonal
For the family of curves represented by the first differential equation the slope of the tangent at any point $C_{1} (x, y)$ is given by ${(\frac{d y}{d x})}_{c_{1}} = \frac{x^{2} + x + 1}{y^{2} + y + 1}$
For the family of curves represented by the second differential the slope of the tangent at any point is given by
${(\frac{d y}{d x})}_{c_{1}} = - \frac{y^{2} + y + 1}{x^{2} + x + 1}$
Clearly, ${(\frac{d y}{d x})}_{c_{1}} \times {(\frac{d y}{d x})}_{c_{1}} = - 1$
Hence, the two curves are orthogonal.

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Similar questions

\frac{d y_{1}}{d x} = \frac{x^{2} + x + 1}{y^{2} + y + 1}

\frac{d y_{2}}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0

y = e^{tan x},

({cos}^{2} x) \cdot y_{2} =

(where y_{2} = \frac{d^{2} y}{d x^{2}}, y_{1} = \frac{d y}{d x})

\frac{d y}{d x} = - {(\frac{x}{y})}^{n}

n = - 1

n = + 1

The differential equation of the family of curves represented by the equation $x^{2} y = a$ , is

Show that the general solution of the differential equation is given by (x + y + 1) = A (1 – x – y – 2xy), where A is parameter

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