मराठी

Write the Differential Equation Obtained Eliminating the Arbitrary Constant C in the Equation Xy = C2. - Mathematics

Advertisements
Advertisements

प्रश्न

Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.

Advertisements

उत्तर

We have,
\[xy = C^2 \]
Differentiating with respect to x, we get
\[x\frac{dy}{dx} + y = 0\]
\[ \Rightarrow x\frac{dy}{dx} = - y\]
\[ \Rightarrow x dy = - y dx\]
\[ \Rightarrow x dy + y dx = 0\]
Hence, x dy + y dx = 0 is the required differential equation .

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 22: Differential Equations - Very Short Answers [पृष्ठ १३८]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 22 Differential Equations
Very Short Answers | Q 6 | पृष्ठ १३८

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

 

Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.


Verify that \[y = e^{m \cos^{- 1} x}\] satisfies the differential equation \[\left( 1 - x^2 \right)\frac{d^2 y}{d x^2} - x\frac{dy}{dx} - m^2 y = 0\]


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

Differential equation Function
\[x\frac{dy}{dx} = y\]
y = ax

\[\frac{dy}{dx} = \log x\]

\[\sin^4 x\frac{dy}{dx} = \cos x\]

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

\[\left( x^3 + x^2 + x + 1 \right)\frac{dy}{dx} = 2 x^2 + x\]

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

\[\left( x - 1 \right)\frac{dy}{dx} = 2 x^3 y\]

xy dy = (y − 1) (x + 1) dx


(1 − x2) dy + xy dx = xy2 dx


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

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

\[\frac{dy}{dx} = 1 + x + y^2 + x y^2\] when y = 0, x = 0

Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]


In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).


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.


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


Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?


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.


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


A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.


Define a differential equation.


Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]


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


Which of the following transformations reduce the differential equation \[\frac{dz}{dx} + \frac{z}{x}\log z = \frac{z}{x^2} \left( \log z \right)^2\] into the form \[\frac{du}{dx} + P\left( x \right) u = Q\left( x \right)\]


Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0


For each of the following differential equations find the particular solution.

(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0


Solve the following differential equation.

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


Solve the following differential equation.

(x2 − y2 ) dx + 2xy dy = 0


Solve the following differential equation.

`dy/dx + 2xy = x`


Solve the differential equation:

dr = a r dθ − θ dr


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c1

∴ e–2y = – 2sin x – 2c1

∴ `square` = c, where c = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×