Advertisements
Advertisements
Question
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Advertisements
Solution
Given equation can be written as
`x^2 "dy"/"dx" - xy = 2cos^2 (y/2x)`, x ≠ 0.
⇒ `(x^2 "dy"/"dx" - xy)/(2cos^2 (y/(2x))` = 1
⇒ `sec^2 (y/(2x))/2 [x^2 "dy"/"dx" - xy]` = 1
Dividing both sides by x3, we get
`sec^2(y/(2x))/2 [(x "dy"/"dx" - y)/x^2] = 1/x^3`
⇒ `"d"/"dx"[tan(y/(2x))] = 1/x^3`
Integrating both sides, we get
`tan(y/(2x)) = (-1)/(2x^2) + "k"`
Substituting x = 1, y = `pi/2`, we get
k = `3/2`
Therefore, `tan(y/(2x)) = -1/(2x^2) + 3/2` is the required solution.
APPEARS IN
RELATED QUESTIONS
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
x cos2 y dx = y cos2 x dy
y ex/y dx = (xex/y + y) dy
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
Experiments show that radium disintegrates at a rate proportional to the amount of radium present at the moment. Its half-life is 1590 years. What percentage will disappear in one year?
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
The slope of the tangent at each point of a curve is equal to the sum of the coordinates of the point. Find the curve that passes through the origin.
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
Show that y = ae2x + be−x is a solution of the differential equation \[\frac{d^2 y}{d x^2} - \frac{dy}{dx} - 2y = 0\]
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
Solve the following differential equation.
`(x + y) dy/dx = 1`
The solution of `dy/dx + x^2/y^2 = 0` is ______
Solve the differential equation:
dr = a r dθ − θ dr
Solve
`dy/dx + 2/ x y = x^2`
Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0
Solve the following differential equation y2dx + (xy + x2) dy = 0
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0
There are n students in a school. If r % among the students are 12 years or younger, which of the following expressions represents the number of students who are older than 12?
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
