मराठी
महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

In each of the following examples, verify that the given function is a solution of the corresponding differential equation. Solution D.E. y = xn x2d2ydx2-n×xdydx+ny=0

Advertisements
Advertisements

प्रश्न

In the following example, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = xn `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0`
बेरीज
Advertisements

उत्तर

y = x n

Differentiating w.r.t. x, we get

`dy/dx = nx^(n-1)`

Again, differentiating w.r.t. x, we get

`(d^2y)/dx^2 = n(n-1) x^(n-2)`

∴  `x^2(d^2y)/dx^2 - nxdy/dx +ny`

= n(n-1)x2xn-2 - nx.nxn-1+ nxn

= n(n-1)xn - n2 xn + nxn

=[n(n-1)-n2+n]xn

= 0

∴ `x^2 (d^2y)/dx^2 - nxdy/dx + ny = 0`

∴ Given function is a solution of the given differential equation.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 8: Differential Equation and Applications - Exercise 8.1 [पृष्ठ १६२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.1 | Q 2.2 | पृष्ठ १६२

संबंधित प्रश्‍न

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

\[\frac{d^4 y}{d x^4} = \left\{ c + \left( \frac{dy}{dx} \right)^2 \right\}^{3/2}\]

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

\[x^2 \left( \frac{d^2 y}{d x^2} \right)^3 + y \left( \frac{dy}{dx} \right)^4 + y^4 = 0\]

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.

 

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


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

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

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

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

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

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.


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

2xy dx + (x2 + 2y2) dy = 0


Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.


If the marginal cost of manufacturing a certain item is given by C' (x) = \[\frac{dC}{dx}\] = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

 

In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]


The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.


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.


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


The solution of the differential equation y1 y3 = y22 is


Solve the following differential equation.

dr + (2r)dθ= 8dθ


Choose the correct alternative.

The solution of `x dy/dx = y` log y is


`xy dy/dx  = x^2 + 2y^2`


Solve the differential equation xdx + 2ydy = 0


Solve the following differential equation y2dx + (xy + x2) dy = 0


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 = – 2c1 

This is general solution.

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

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×