Advertisements
Advertisements
प्रश्न
Find the general solution of the differential equation `(1 + y^2) + (x - "e"^(tan - 1y)) "dy"/"dx"` = 0.
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.
APPEARS IN
संबंधित प्रश्न
Solve the differential equation: `x+ydy/dx=sec(x^2+y^2)` Also find the particular solution if x = y = 0.
Solve the differential equation `dy/dx=(y+sqrt(x^2+y^2))/x`
Form the differential equation of the family of circles in the second quadrant and touching the coordinate axes.
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 log(dy/dx)= 3x + 4y, given that y = 0 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 = ex + 1 : y″ – y′ = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = Ax : xy′ = y (x ≠ 0)
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
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 `dy/dx + y cot x = 4xcosec x(x != 0)`, given that y = 0 when `x = pi/2.`
if `y = sin^(-1) (6xsqrt(1-9x^2))`, `1/(3sqrt2) < x < 1/(3sqrt2)` then find `(dy)/(dx)`
The solution of the differential equation \[\frac{dy}{dx} = 1 + x + y^2 + x y^2 , y\left( 0 \right) = 0\] is
Solution of the differential equation \[\frac{dy}{dx} + \frac{y}{x}=\sin x\] is
Solve the differential equation (x2 − yx2) dy + (y2 + x2y2) dx = 0, given that y = 1, when x = 1.
The solution of the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\phi\left( \frac{y}{x} \right)}{\phi'\left( \frac{y}{x} \right)}\] is
\[\frac{dy}{dx} = \left( x + y \right)^2\]
\[\left( 1 + y^2 \right) + \left( x - e^{- \tan^{- 1} y} \right)\frac{dy}{dx} = 0\]
For the following differential equation, find the general solution:- \[\frac{dy}{dx} = \frac{1 - \cos x}{1 + \cos x}\]
For the following differential equation, find a particular solution satisfying the given condition:
\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y = 0\text{ when }x = 2\]
Solve the following differential equation:- \[x \cos\left( \frac{y}{x} \right)\frac{dy}{dx} = y \cos\left( \frac{y}{x} \right) + x\]
Solve the following differential equation:-
\[\frac{dy}{dx} - y = \cos x\]
Solve the following differential equation:-
\[\frac{dy}{dx} + \frac{y}{x} = x^2\]
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 a particular solution of the following differential equation:- (x + y) dy + (x − y) dx = 0; y = 1 when x = 1
Find a particular solution of the following differential equation:- x2 dy + (xy + y2) dx = 0; y = 1 when x = 1
Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx, x ≠ 0.
Solution of the differential equation `"dx"/x + "dy"/y` = 0 is ______.
The general solution of the differential equation `"dy"/"dx" + y sec x` = tan x is y(secx – tanx) = secx – tanx + x + k.
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`.
Find the general solution of y2dx + (x2 – xy + y2) dy = 0.
If y = e–x (Acosx + Bsinx), then y is a solution of ______.
The number of solutions of `("d"y)/("d"x) = (y + 1)/(x - 1)` when y (1) = 2 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) + (1 + y^2)/(1 + x^2)` is ______.
The number of arbitrary constants in the general solution of a differential equation of order three is ______.
General solution of `("d"y)/("d"x) + y` = sinx is ______.
The integrating factor of `("d"y)/("d"x) + y = (1 + y)/x` is ______.
Polio drops are delivered to 50 K children in a district. The rate at which polio drops are given is directly proportional to the number of children who have not been administered the drops. By the end of 2nd week half the children have been given the polio drops. How many will have been given the drops by the end of 3rd week can be estimated using the solution to the differential equation `"dy"/"dx" = "k"(50 - "y")` where x denotes the number of weeks and y the number of children who have been given the drops.
The value of c in the particular solution given that y(0) = 0 and k = 0.049 is ______.
The differential equation of all parabolas that have origin as vertex and y-axis as axis of symmetry is ______.
