मराठी

Find the general solution of (1 + tany)(dx – dy) + 2xdy = 0. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the general solution of (1 + tany)(dx – dy) + 2xdy = 0.

बेरीज
Advertisements

उत्तर

Given that: (1 + tan y)(dx – dy) + 2xdy = 0

⇒ (1 + tan y)dx – (1 + tan y)dy + 2xdy = 0

⇒ (1 + tan y)dx – (1 + tan y – 2x)dy = 0

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

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

⇒ `"dx"/"dy" = 1 - (2x)/(1 + tan y)`

⇒ `"dx"/"dy" + (2x)/(1 + tan y)` = 1

Here, P = `2/(1 + tan y)` and Q = 1

Integrating factor I.F.

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

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

= `"e"^(int (siny + cosy - siny + cosy)/((siny + cosy)) "dy"`

= `"e"^(int(1 + (cosy - siny)/(siny + cosy))"d"y)`

= `"e"^(int 1."d"y) . "e"^(int(cosy - siny)/(siny + cosy)"d"y)`

= `"e"^y . "e"^(log(siny + cosy)`

= `"e"^y . (siny + cos y)`

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

⇒ `x . "e"^y (siny + cosy) = int 1 . "e"^y (siny + cosy)"d"y + "c"`

⇒ `x . "e"^y )siny + cosy) = "e"^y . sin y + "c"`  .....`[because int x^x "f"(x) + "f'"(x)]"d"x = "e"^x "f"(x) + "c"]`

⇒ `x(siny + cos y) = sin y + "c" . "e"^-y`

Hence, the required solution is `x(siny + cos y) = sin y + "c" . "e"^-y`.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 9: Differential Equations - Exercise [पृष्ठ १९४]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
पाठ 9 Differential Equations
Exercise | Q 26 | पृष्ठ १९४

व्हिडिओ ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्‍न

The differential equation of `y=c/x+c^2` is :

(a)`x^4(dy/dx)^2-xdy/dx=y`

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

(c)`x^3(dy/dx)^2+xdy/dx=y`

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


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


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


If y = P eax + Q ebx, show that

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


Find the particular solution of the differential equation dy/dx=1 + x + y + xy, given that y = 0 when x = 1.


Solve the differential equation `dy/dx -y =e^x`


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

y = x2 + 2x + C  :  y′ – 2x – 2 = 0


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)`


Write the order of the differential equation associated with the primitive y = C1 + C2 ex + C3 e−2x + C4, where C1, C2, C3, C4 are arbitrary constants.


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


The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is


The solution of the differential equation \[2x\frac{dy}{dx} - y = 3\] represents


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


x (e2y − 1) dy + (x2 − 1) ey dx = 0


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


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


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


(x2 + 1) dy + (2y − 1) dx = 0


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


`x cos x(dy)/(dx)+y(x sin x + cos x)=1`


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


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Solve the following differential equation:-

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


Find a particular solution of the following differential equation:- \[\left( 1 + x^2 \right)\frac{dy}{dx} + 2xy = \frac{1}{1 + x^2}; y = 0,\text{ when }x = 1\]


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


The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is ______.


The general solution of the differential equation `"dy"/"dx" + y sec x` = tan x is y(secx – tanx) = secx – tanx + x + k.


Solve the differential equation dy = cosx(2 – y cosecx) dx given that y = 2 when x = `pi/2`


Solution of differential equation xdy – ydx = 0 represents : ______.


Integrating factor of the differential equation `("d"y)/("d"x) + y tanx - secx` = 0 is ______.


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


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


General solution of the differential equation of the type `("d"x)/("d"x) + "P"_1x = "Q"_1` is given by ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×