मराठी

The general solution of the differential equation ddedydx=ex22+xy is ______. - Mathematics

Advertisements
Advertisements

प्रश्न

The general solution of the differential equation `("d"y)/("d"x) = "e"^(x^2/2) + xy` is ______.

पर्याय

  • y = `"ce"^((-x^2)/2`

  • y = `"ce"^((x^2)/2`

  • y = `(x + "c")"e"^((x^2)/2`

  • y = `("c" - x)"e"^((x^2)/2`

MCQ
रिकाम्या जागा भरा

उत्तर

The general solution of the differential equation `("d"y)/("d"x) = "e"^(x^2/2) + xy` is y = `(x + "c")"e"^((x^2)/2`.

Explanation:

The given differential equation is `("d"y)/("d"x) = "e"^(x^2/2) + xy`

⇒ `("d"y)/("d"x) - xy = "e"^((x^2)/2`

Since it is linear differential equation

Where P = –x and Q = `"e"^((x^2)/2`

∴ Integrating factor I.F. = `"e"^(int Pdx)`

= `"e"^(int -x  "d"x)`

= `"e"^(- x^2/2)`

So, the solution is `y xx "I"."F". = int "Q" xx "I"."F".  "d"x + "c"`

⇒ `y xx "e"^( x^2/2) = int "e"^(x^2/2) "e"^(- x^2/2)  "d"x + "c"`

⇒ `y xx "e"^(- x^2/2) = int "e"^0  "d"x + "c"`

⇒ `y xx "e"^(- x^2/2) = int 1 . "d"x + "c"`

⇒ `y xx "e"^(- x^2/2) = x + "c"`

∴ y = `(x + "c")"e"^(x^2/2)`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Equations - Exercise [पृष्ठ १९९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 9 Differential Equations
Exercise | Q 63 | पृष्ठ १९९

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

If   `y=sqrt(sinx+sqrt(sinx+sqrt(sinx+..... oo))),` then show that `dy/dx=cosx/(2y-1)`


The solution of the differential equation dy/dx = sec x – y tan x is:

(A) y sec x = tan x + c

(B) y sec x + tan x = c

(C) sec x = y tan x + c

(D) sec x + y tan x = c


Solve the differential equation:  `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 0.


If x = Φ(t) differentiable function of ‘ t ' then prove that `int f(x) dx=intf[phi(t)]phi'(t)dt`


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


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


The number of arbitrary constants in the general solution of a differential equation of fourth order are ______.


How many arbitrary constants are there in the general solution of the differential equation of order 3.


Solve the differential equation (x2 − yx2) dy + (y2 + x2y2) dx = 0, given that y = 1, when x = 1.


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


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


(x3 − 2y3) dx + 3x2 y dy = 0


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


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


Solve the following differential equation:-

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


Solve the following differential equation:-

(1 + x2) dy + 2xy dx = cot x dx


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 ______.


Solution of `("d"y)/("d"x) - y` = 1, y(0) = 1 is given by ______.


tan–1x + tan–1y = c is the general solution of the differential equation ______.


The differential equation for which y = acosx + bsinx is a solution, is ______.


The number of arbitrary constants in the general solution of a differential equation of order three is ______.


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


Find a particular solution satisfying the given condition `- cos((dy)/(dx)) = a, (a ∈ R), y` = 1 when `x` = 0


If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×