हिंदी

Find the general solution of the differential equation edydx(1+y2)+(x-etan-1y)dydx = 0.

Advertisements
Advertisements

प्रश्न

Find the general solution of the differential equation `(1 + y^2) + (x - "e"^(tan - 1y)) "dy"/"dx"` = 0.

योग
Advertisements

उत्तर

Given equation is `(1 + y^2) + (x - "e"^(tan^(-1) y)) "dy"/"dx"` = 0

⇒ `(x - "e"^(tan^-1y)) "dy"/"dx" = -(1 + y^2)`

⇒ `"dy"/"dx" = (-(1 + y^2))/(x - "e"^(tan^-1 y))`

⇒ `"dx"/"dy" = (x - "e"^(tan^-1y))/(-(1 + y^2))`

⇒ `"dx"/"dy" = - x/((1 + y^2)) + ("e"^(tan^-1y))/(1 + y^2)` 

⇒ `"dx"/"dy" + x/((1 + y^2)) = ("e"^(tan^-1 y))/(1 + y^2)`

Here, P = `1/(1 + y^2)` and Q = `("e"^(tan^-1 y))/(1 + y^2)`

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

= `"e"^(int 1/(1 + y^2) "d"y)`

= `"e"^(tan^-1 y)`

∴ Solution is `x . "I"."F". = int "Q". "I"."F".  "d"y + "c"`

⇒ `x . "e"^(tan^-1 y) = int ("e"^(tan^-1 y))/(1 + y^2) * "e"^(tan^-1 y) "dy" + "c"`

Put `"e"^(tan^-1 y)` = t

∴ `"e"^(tan^-1 y) * 1/(1 + y^2) "dy"` = dt

∴ `x . "e"^(tan^-1 y) = int "t" . "dt" + "c"`

⇒ `x . "e"^(tan^-1 y) = 1/2 "t"^2 + "c"`

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

⇒ x = `1/2 ("e"^(tan^-1 y)) + "c"/("e"^(tan^-1 y))`

⇒ 2x = `"e"^(tan^-1 y) + (2"c")/("e"^(tan^-1 y)`

⇒ `2x . "e"^(tan^-1 y) = ("e"^(tan^-1y))^2 + 2"c"`

Hence, this is the required general solution.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Exercise [पृष्ठ १९४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 9 Differential Equations
Exercise | Q 17 | पृष्ठ १९४

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

Find the particular solution of the differential equation

(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.


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

y – cos y = x :  (y sin y + cos y + x) y′ = 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


Find the general solution of the differential equation `dy/dx + sqrt((1-y^2)/(1-x^2)) = 0.`


Find a particular solution of the differential equation `dy/dx + y cot x = 4xcosec x(x != 0)`, given that y = 0 when `x = pi/2.`


Find `(dy)/(dx)` at x = 1, y = `pi/4` if `sin^2 y + cos xy = K`


Find the particular solution of the differential equation

`tan x * (dy)/(dx) = 2x tan x + x^2 - y`; `(tan x != 0)` given that y = 0 when `x = pi/2`


The solution of the differential equation \[x\frac{dy}{dx} = y + x \tan\frac{y}{x}\], is


The number of arbitrary constants in the general solution of differential equation of fourth order is


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


x (e2y − 1) dy + (x2 − 1) ey dx = 0


(x3 − 2y3) dx + 3x2 y dy = 0


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


`2 cos x(dy)/(dx)+4y sin x = sin 2x," given that "y = 0" when "x = pi/3.`


For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \left( 1 + x^2 \right)\left( 1 + y^2 \right)\]


For the following differential equation, find the general solution:- \[\frac{dy}{dx} + y = 1\]


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

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


For the following differential equation, find a particular solution satisfying the given condition:- \[\frac{dy}{dx} = y \tan x, y = 1\text{ when }x = 0\]


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Solve the following differential equation:-

(1 + x2) dy + 2xy dx = cot x dx


Find the equation of a curve passing through the point (0, 1). If the slope of the tangent to the curve at any point (x, y) is equal to the sum of the x-coordinate and the product of the x-coordinate and y-coordinate of that point.


Solution of the differential equation `"dx"/x + "dy"/y` = 0 is ______.


Find the general solution of y2dx + (x2 – xy + y2) dy = 0.


Solve the differential equation (1 + y2) tan–1xdx + 2y(1 + x2)dy = 0.


Solve: `y + "d"/("d"x) (xy) = x(sinx + logx)`


The solution of the differential equation `("d"y)/("d"x) + (1 + y^2)/(1 + x^2)` is ______.


y = aemx+ be–mx satisfies which of the following differential equation?


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


The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is ______.


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


The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is ______.


General solution of `("d"y)/("d"x) + y` = sinx is ______.


The integrating factor of `("d"y)/("d"x) + y = (1 + y)/x` is ______.


The solution of the differential equation `("d"y)/("d"x) = (x + 2y)/x` is x + y = kx2.


Find a particular solution, satisfying the condition `(dy)/(dx) = y tan x ; y = 1` when `x = 0`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×