मराठी

Find the the differential equation for all the straight lines, which are at a unit distance from the origin. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the the differential equation for all the straight lines, which are at a unit distance from the origin.

उत्तर

The general equation of a line that is at unit distance from the origin is given by

`xcosα+ysinα=1      .....(i)`



Differentiating (i) w.r.t. x, we get

`cosα+dy/dxsinα=0`

`⇒cotα=−dy/dx           .....(ii)`

Dividing (i) by sinα, we get

`x cosα/sinα+ysinα/sinα=1/sinα `

`⇒xcotα+y=cosecα`

`⇒xcotα+y=sqrt(1+cot^2α)      .....(iii)`


 

Putting the value of (ii) in (iii), we get

`x(−dy/dx)+y=sqrt(1+(−dy/dx)^2)  .....(iv)`

Squaring (iv), we get

`(−xdy/dx+y)^2=(sqrt(1+(dy/dx)^2))^2`

`(x^2−1)(dy/dx)^2−2xydy/dx+y^2−1=0`

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2014-2015 (March) Patna Set 2

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

For the differential equation, find the general solution:

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


For the differential equation, find the general solution:

`(x + 3y^2) dy/dx = y(y > 0)`


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

\[\left( 2x - 10 y^3 \right)\frac{dy}{dx} + y = 0\]

(x + tan y) dy = sin 2y dx


dx + xdy = e−y sec2 y dy


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

\[\frac{dy}{dx} + 2y = x e^{4x}\]

Find the general solution of the differential equation \[x\frac{dy}{dx} + 2y = x^2\]


Find the general solution of the differential equation \[\frac{dy}{dx} - y = \cos x\]


Solve the following differential equation: \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\] .


Solve the following differential equation:-
\[\left( 1 + x^2 \right)\frac{dy}{dx} - 2xy = \left( x^2 + 2 \right)\left( x^2 + 1 \right)\]


Find the integerating factor of the differential equation `xdy/dx - 2y = 2x^2` . 


Solve the differential equation: (1 +x) dy + 2xy dx = cot x dx 


If f(x) = x + 1, find `"d"/"dx"("fof") ("x")`


Solve the differential equation: `(1 + x^2) dy/dx + 2xy - 4x^2 = 0,` subject to the initial condition y(0) = 0.


Solve the following differential equation:

`"dy"/"dx" + "y"/"x" = "x"^3 - 3`


Solve the following differential equation:

`"dy"/"dx" + "y" * sec "x" = tan "x"`


Solve the following differential equation dr + (2r cot θ + sin 2θ) dθ = 0.


Solve the following differential equation:

y dx + (x - y2) dy = 0


Solve the following differential equation:

`(1 - "x"^2) "dy"/"dx" + "2xy" = "x"(1 - "x"^2)^(1/2)`


Find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it.


Find the equation of the curve passing through the point `(3/sqrt2, sqrt2)` having a slope of the tangent to the curve at any point (x, y) is -`"4x"/"9y"`.


If y = y(x) is the solution of the differential equation, `(dy)/(dx) + 2ytanx = sinx, y(π/3)` = 0, then the maximum value of the function y (x) over R is equal to ______.


Let y = f(x) be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If f(x) satisfies xf'(x) = x2 + f(x) – 2, then the area bounded by f(x) with x-axis between ordinates x = 0 and x = 3 is equal to ______.


Find the general solution of the differential equation:

`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`


Solve:

`xsinx dy/dx + (xcosx + sinx)y` = sin x


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×