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 - Mathematics and Statistics

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

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] 12 Standard HSC Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.1 | Q 2.2 | पृष्ठ १६२

संबंधित प्रश्‍न

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

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

Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 1\] Function y = sin x + cos x


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

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

\[\frac{dy}{dx} = \frac{x\left( 2 \log x + 1 \right)}{\sin y + y \cos y}\]

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

Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.


2xy dx + (x2 + 2y2) dy = 0


Solve the following initial value problem:-

\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]


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).


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.


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

y = ex + 1            y'' − y' = 0


Determine the order and degree of the following differential equations.

Solution D.E.
ax2 + by2 = 5 `xy(d^2y)/dx^2+ x(dy/dx)^2 = y dy/dx`

Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


State whether the following statement is True or False:

The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x 


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


Solve the following differential equation 

sec2 x tan y dx + sec2 y tan x dy = 0

Solution: sec2 x tan y dx + sec2 y tan x dy = 0

∴ `(sec^2x)/tanx  "d"x + square` = 0

Integrating, we get

`square + int (sec^2y)/tany  "d"y` = log c

Each of these integral is of the type

`int ("f'"(x))/("f"(x))  "d"x` = log |f(x)| + log c

∴ the general solution is

`square + log |tan y|` = log c

∴ log |tan x . tan y| = log c

`square`

This is the general solution.


The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:


Solve the differential equation

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


The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×