हिंदी

Y = x is a particular solution of the differential equation dddydxd2ydx2-x2dydx+xy = x. - Mathematics

Advertisements
Advertisements

प्रश्न

y = x is a particular solution of the differential equation `("d"^2y)/("d"x^2) - x^2 "dy"/"dx" + xy` = x.

विकल्प

  • True

  • False

MCQ
सत्य या असत्य

उत्तर

This statement is False.

Explanation:

Because y = x does not satisfy the given differential equation.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Solved Examples [पृष्ठ १९२]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 9 Differential Equations
Solved Examples | Q 23. (x) | पृष्ठ १९२

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

The differential equation of the family of curves y=c1ex+c2e-x is......

(a)`(d^2y)/dx^2+y=0`

(b)`(d^2y)/dx^2-y=0`

(c)`(d^2y)/dx^2+1=0`

(d)`(d^2y)/dx^2-1=0`


Find the differential equation representing the curve y = cx + c2.


Find the particular solution of the differential equation `(1+x^2)dy/dx=(e^(mtan^-1 x)-y)` , give that y=1 when x=0.


Find the particular solution of the differential equation log(dy/dx)= 3x + 4y, given that y = 0 when x = 0.


Find the particular solution of the differential equation x (1 + y2) dx – y (1 + x2) dy = 0, given that y = 1 when x = 0.


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = cos x + C : y′ + sin x = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

`y sqrt(1 + x^2) : y' = (xy)/(1+x^2)`


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x sin x : xy' = `y + x  sqrt (x^2 - y^2)`  (x ≠ 0 and x > y or x < -y)


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

x + y = tan–1y   :   y2 y′ + y2 + 1 = 0


Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

`y = sqrt(a^2 - x^2 )  x in (-a,a) : x + y  dy/dx = 0(y != 0)`


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


Solve the differential equation:

`e^(x/y)(1-x/y) + (1 + e^(x/y)) dx/dy = 0` when x = 0, y = 1


The solution of the differential equation \[\frac{dy}{dx} + \frac{2y}{x} = 0\] with y(1) = 1 is given by


Which of the following differential equations has y = x as one of its particular solution?


\[\frac{dy}{dx} - y \tan x = e^x \sec x\]


\[\frac{dy}{dx} + 2y = \sin 3x\]


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


`(dy)/(dx)+ y tan x = x^n cos x, n ne− 1`


For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sqrt{4 - y^2}, - 2 < y < 2\]


For the following differential equation, find a particular solution satisfying the given condition:

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


Solve the following differential equation:- `y dx + x log  (y)/(x)dy-2x dy=0`


Solve the following differential equation:-

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


Find a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1


Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]


x + y = tan–1y is a solution of the differential equation `y^2 "dy"/"dx" + y^2 + 1` = 0.


If y(t) is a solution of `(1 + "t")"dy"/"dt" - "t"y` = 1 and y(0) = – 1, then show that y(1) = `-1/2`.


Integrating factor of the differential equation `cosx ("d"y)/("d"x) + ysinx` = 1 is ______.


The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.


Number of arbitrary constants in the particular solution of a differential equation of order two is two.


Find the general solution of the differential equation:

`log((dy)/(dx)) = ax + by`.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×