English

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

Advertisements
Advertisements

Question

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

Sum
Advertisements

Solution

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
  Is there an error in this question or solution?
Chapter 9: Differential Equations - Exercise [Page 194]

APPEARS IN

NCERT Exemplar Mathematics Exemplar [English] Class 12
Chapter 9 Differential Equations
Exercise | Q 26 | Page 194

RELATED QUESTIONS

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`


Solve : 3ex tanydx + (1 +ex) sec2 ydy = 0

Also, find the particular solution when x = 0 and y = π.


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)


The number of arbitrary constants in the particular solution of a differential equation of third order are ______.


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


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


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


Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is


If m and n are the order and degree of the differential equation \[\left( y_2 \right)^5 + \frac{4 \left( y_2 \right)^3}{y_3} + y_3 = x^2 - 1\], then


The solution of the differential equation \[\frac{dy}{dx} + 1 = e^{x + y}\], is


The solution of the differential equation \[\frac{dy}{dx} - ky = 0, y\left( 0 \right) = 1\] approaches to zero when x → ∞, if


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


The general solution of the differential equation \[\frac{y dx - x dy}{y} = 0\], is


The general solution of a differential equation of the type \[\frac{dx}{dy} + P_1 x = Q_1\] is


Find the general solution of the differential equation \[x \cos \left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x .\]


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

 

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


(x + y − 1) dy = (x + y) dx


(1 + y + x2 y) dx + (x + x3) dy = 0


\[\frac{dy}{dx} + y = 4x\]


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


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


Find the general solution of the differential equation \[\frac{dy}{dx} = \frac{x + 1}{2 - y}, y \neq 2\]


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 the general solution:- \[\frac{dy}{dx} = \sin^{- 1} x\]


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


Solve the following differential equation:-

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


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


Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution.


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


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


The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.


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


The general solution of `("d"y)/("d"x) = 2x"e"^(x^2 - y)` is ______.


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


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


The differential equation of all parabolas that have origin as vertex and y-axis as axis of symmetry is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×