Advertisements
Advertisements
Question
Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).
Advertisements
Solution

Portion of the x-axis cut off between the origin and tangent at a point \[= x - y \hspace{0.167em} \hspace{0.167em} \frac{dx}{dy} = OT\]
It is given, OT = 2x
\[\begin{array}{l}\therefore \hspace{0.167em} \hspace{0.167em} x - y \hspace{0.167em} \hspace{0.167em} \frac{dx}{dy} = 2x \\ - x = y\frac{dx}{dy} \\ - \int\frac{dx}{x} = \int\frac{dy}{y} \\ \therefore \hspace{0.167em} \hspace{0.167em} xy = k\end{array}\]
Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2
APPEARS IN
RELATED QUESTIONS
Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]
Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\] satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]
C' (x) = 2 + 0.15 x ; C(0) = 100
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
The differential equation
\[\frac{dy}{dx} + Py = Q y^n , n > 2\] can be reduced to linear form by substituting
Form the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
The solution of `dy/dx + x^2/y^2 = 0` is ______
Solve
`dy/dx + 2/ x y = x^2`
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.
Solve: ydx – xdy = x2ydx.
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
Solve the differential equation
`y (dy)/(dx) + x` = 0
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
