Advertisements
Advertisements
Question
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Advertisements
Solution
The equation of the family of hyperbolas having the centre at the origin and foci on the x-axis is \[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1............(1)\]
where \[a\text{ and }b\] are parameters.
As this equation contains two parameters, we shall get a second-order differential equation.
Differentiating equation (1) with respect to x, we get
\[\frac{2x}{a^2} - \frac{2y}{b^2}\frac{dy}{dx} =0..........(2)\]
Differentiating equation (2) with respect to x, we get
\[\frac{2}{a^2} - \frac{2}{b^2}\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] = 0\]
\[ \Rightarrow \frac{1}{a^2} = \frac{1}{b^2}\left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right]\]
\[ \Rightarrow \frac{b^2}{a^2} = \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right] \left......( 3 \right)\]
Now, from equation (2), we get
\[\frac{2x}{a^2} = \frac{2y}{b^2}\frac{dy}{dx}\]
\[ \Rightarrow \frac{b^2}{a^2} = \frac{y}{x}\frac{dy}{dx} ........(4)\]
From (3) and (4), we get
\[\frac{y}{x}\frac{dy}{dx} = \left[ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right]\]
\[ \Rightarrow y\frac{dy}{dx} = xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 \]
\[ \Rightarrow xy\frac{d^2 y}{d x^2} + x \left( \frac{dy}{dx} \right)^2 - y\frac{dy}{dx} = 0\]
It is the required differential equation.
APPEARS IN
RELATED QUESTIONS
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
Find the curve for which the intercept cut-off by a tangent on x-axis is equal to four times the ordinate of the point of contact.
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
The solution of the differential equation y1 y3 = y22 is
The differential equation satisfied by ax2 + by2 = 1 is
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = xn | `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0` |
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
Solve the following differential equation.
`dy /dx +(x-2 y)/ (2x- y)= 0`
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
`xy dy/dx = x^2 + 2y^2`
Solve the following differential equation y2dx + (xy + x2) dy = 0
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.
