Advertisements
Advertisements
Question
Integrating factor of the differential equation `"dy"/"dx" - y` = cos x is ex.
Options
True
False
Solution
This statement is False.
Explanation:
Because I.F = `"e"^(int -1 "d"x)`
= e–x.
APPEARS IN
RELATED QUESTIONS
Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.
Form the differential equation representing the family of ellipses having centre at the origin and foci on x-axis.
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
Differential equation \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 1\] Function y = sin x + cos x
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\]
(x + y) (dx − dy) = dx + dy
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y \tan x = \sin x; y = 0\text{ when }x = \frac{\pi}{3}\]
The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
The differential equation satisfied by ax2 + by2 = 1 is
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
Solve the following differential equation.
dr + (2r)dθ= 8dθ
The solution of `dy/ dx` = 1 is ______
The solution of `dy/dx + x^2/y^2 = 0` is ______
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
Solve the differential equation `("d"y)/("d"x) + y` = e−x
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
