मराठी

Find the differential equation representing the family of curves v=A/r+ B, where A and B are arbitrary constants. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the differential equation representing the family of curves v=A/r+ B, where A and B are arbitrary constants.

Advertisements

उत्तर

 

The equation of the family of curves is v=A/r+B, where A and B are arbitrary constants.

We have

v=Ar+B

Differentiating both sides with respect to r, we get

`(dv)/(dr)=-A/r^2+0`

`⇒r^2(dv)/(dr)=−A`

Again, differentiating both sides with respect to r, we get

`r^2xx(d^2v)/(d^2r)+2rxx(dv)/(dr)=0`

`⇒r(d^2v)/(d^2r)+2(dv)/(dr)=0`

This is the differential equation representing the family of the given curves

 
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
2014-2015 (March) Delhi Set 1

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Find the differential equation of the family of lines passing through the origin.


Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

`x/a + y/b = 1`


Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = e2x (a + bx)


Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.

y = ex (a cos x + b sin x)


Find a particular solution of the differential equation (x - y) (dx + dy) = dx - dy, given that y = -1, when x = 0. (Hint: put x - y = t)


The general solution of the differential equation `(y dx - x dy)/y = 0` is ______.


Find the differential equation representing the family of curves `y = ae^(bx + 5)`. where a and b are arbitrary constants.


Find the differential equation corresponding to y = ae2x + be3x + cex where abc are arbitrary constants.


Show that the differential equation of all parabolas which have their axes parallel to y-axis is \[\frac{d^3 y}{d x^3} = 0.\]


\[\frac{dy}{dx} = \sin^3 x \cos^4 x + x\sqrt{x + 1}\]


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


\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]


\[(\tan^2 x + 2\tan x + 5)\frac{dy}{dx} = 2(1+\tan x)\sec^2x\]


\[\frac{dy}{dx} = \sin^3 x \cos^2 x + x e^x\]


tan y dx + tan x dy = 0


x cos2 y dx = y cos2 x dy


cosec x (log y) dy + x2y dx = 0


Find the general solution of the differential equation `"dy"/"dx" = y/x`.


Solve the differential equation:

cosec3 x dy − cosec y dx = 0


Find the general solution of the following differential equation:

`x (dy)/(dx) = y - xsin(y/x)`


The general solution of the differential equation `(dy)/(dx) + x/y` = 0 is


General solution of tan 5θ = cot 2θ is


The general solution of the differential equation `(dy)/(dx) = e^(x + y)` is


The general solution of the differential equation `(ydx - xdy)/y` = 0


Find the general solution of differential equation `(dy)/(dx) = (1 - cosx)/(1 + cosx)`


Solve the differential equation: y dx + (x – y2)dy = 0


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×