Advertisements
Advertisements
प्रश्न
tan–1x + tan–1y = c is the general solution of the differential equation ______.
विकल्प
`("d"y)/("d"x) = (1 + y^2)/(1 + x^2)`
`("d"y)/("d"x) = (1 + x^2)/(1 + y^2)`
(1 + x2)dy + (1 + y2)dx = 0
(1 + x2)dx + (1 + y2)dy = 0
Advertisements
उत्तर
tan–1x + tan–1y = c is the general solution of the differential equation (1 + x2)dy + (1 + y2)dx = 0.
Explanation:
Given equation is tan–1x + tan–1y = c
Differentiating w.r.t. x, we have
`1/(1 + x^2) + 1/(1 + y^2) * ("d"y)/("d"x)` = 0
⇒ `(1/(1 + y^2)) ("d"y)/("d"x) = -(1/(1 + x^2))`
⇒ `("d"y)/("d"x) = -((1 + y^2)/(1 + x^2))`
⇒ (1 + x2)dy = – (1 + y2)dx
⇒ (1 + x2)dy + (1 + y2)dx = 0.
APPEARS IN
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.
Find the general solution of the following differential equation :
`(1+y^2)+(x-e^(tan^(-1)y))dy/dx= 0`
Find the particular solution of the differential equation `dy/dx=(xy)/(x^2+y^2)` given that y = 1, when x = 0.
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 = cos x + C : y′ + sin x = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
xy = log y + C : `y' = (y^2)/(1 - xy) (xy != 1)`
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)`
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
The solution of x2 + y2 \[\frac{dy}{dx}\]= 4, is
The solution of the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + 1 + y^2 = 0\], is
The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is
The general solution of the differential equation ex dy + (y ex + 2x) dx = 0 is
\[\frac{dy}{dx} + 1 = e^{x + y}\]
\[\frac{dy}{dx} - y \cot x = cosec\ x\]
(x2 + 1) dy + (2y − 1) dx = 0
`(2ax+x^2)(dy)/(dx)=a^2+2ax`
`x cos x(dy)/(dx)+y(x sin x + cos x)=1`
\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \sin^{- 1} x\]
Solve the following differential equation:- \[\left( x - y \right)\frac{dy}{dx} = x + 2y\]
Solve the following differential equation:- `y dx + x log (y)/(x)dy-2x dy=0`
Solve the following differential equation:-
\[\frac{dy}{dx} + \frac{y}{x} = x^2\]
Solve the following differential equation:-
\[\left( x + y \right)\frac{dy}{dx} = 1\]
Find a particular solution of the following differential equation:- x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
The number of arbitrary constants in a particular solution of the differential equation tan x dx + tan y dy = 0 is ______.
x + y = tan–1y is a solution of the differential equation `y^2 "dy"/"dx" + y^2 + 1` = 0.
Find the general solution of `(x + 2y^3) "dy"/"dx"` = y
Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution.
Solution of the differential equation tany sec2xdx + tanx sec2ydy = 0 is ______.
The solution of the differential equation cosx siny dx + sinx cosy dy = 0 is ______.
The solution of `x ("d"y)/("d"x) + y` = ex is ______.
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
The solution of the differential equation `("d"y)/("d"x) + (2xy)/(1 + x^2) = 1/(1 + x^2)^2` is ______.
The solution of differential equation coty dx = xdy is ______.
The integrating factor of `("d"y)/("d"x) + y = (1 + y)/x` is ______.
Which of the following differential equations has `y = x` as one of its particular solution?
Find the general solution of the differential equation:
`(dy)/(dx) = (3e^(2x) + 3e^(4x))/(e^x + e^-x)`
