हिंदी

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).Find Th

Advertisements
Advertisements

प्रश्न

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

उत्तर


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 = 2
\[\begin{array}{l}\therefore \hspace{0.167em} \hspace{0.167em} x - y \hspace{0.167em} \hspace{0.167em} \frac{dx}{dy} = 2x \\ - x = y\frac{dx}{dy} \\ - \int\frac{dx}{x} = \int\frac{dy}{y} \\ \therefore \hspace{0.167em} \hspace{0.167em} xy = k\end{array}\]
Since the curve passes through the point (1, 2)
⇒ at x = 1 ⇒ y = 2
∴ k = 2
∴ xy = 2

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 21: Differential Equations - Exercise 22.11 [पृष्ठ १३५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 21 Differential Equations
Exercise 22.11 | Q 19 | पृष्ठ १३५

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].

 


Verify that y = cx + 2c2 is a solution of the differential equation 

\[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0\].

Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]


Show that y = e−x + ax + b is solution of the differential equation\[e^x \frac{d^2 y}{d x^2} = 1\]

 


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x + y\frac{dy}{dx} = 0\]
\[y = \pm \sqrt{a^2 - x^2}\]

\[x\frac{dy}{dx} + 1 = 0 ; y \left( - 1 \right) = 0\]

xy dy = (y − 1) (x + 1) dx


y (1 + ex) dy = (y + 1) ex dx


\[\frac{dy}{dx} = 1 - x + y - xy\]

\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

\[2x\frac{dy}{dx} = 5y, y\left( 1 \right) = 1\]

\[\frac{dy}{dx} = 2xy, y\left( 0 \right) = 1\]

\[\frac{dy}{dx} = \frac{\left( x - y \right) + 3}{2\left( x - y \right) + 5}\]

(x2 − y2) dx − 2xy dy = 0


\[\frac{dy}{dx} = \frac{x + y}{x - y}\]

\[x^2 \frac{dy}{dx} = x^2 - 2 y^2 + xy\]

\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]


(x + 2y) dx − (2x − y) dy = 0


Solve the following initial value problem:-

\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]


The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.


The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).


Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.


If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is


Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

`y=sqrt(a^2-x^2)`              `x+y(dy/dx)=0`


Solve the following differential equation.

x2y dx − (x3 + y3) dy = 0


Solve the following differential equation.

`(x + y) dy/dx = 1`


 `dy/dx = log x`


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0


For the differential equation, find the particular solution

`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0


Solve the following differential equation

`y log y ("d"x)/("d"y) + x` = log y


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.


Solution of `x("d"y)/("d"x) = y + x tan  y/x` is `sin(y/x)` = cx


Why is the equation \[x\frac{dy}{dx} + y = 0\] classified as a differential equation?


Which of the following is an example of an ordinary differential equation?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×