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
Solve the equation for x: `sin^(-1) 5/x + sin^(-1) 12/x = π/2, x ≠ 0`
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]
Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
Differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 2\] Function y = xex + ex
tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y)
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after `t` seconds.
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.
The population of a city increases at a rate proportional to the number of inhabitants present at any time t. If the population of the city was 200000 in 1990 and 250000 in 2000, what will be the population in 2010?
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
For the following differential equation find the particular solution.
`dy/ dx = (4x + y + 1),
when y = 1, x = 0
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
`x^2 dy/dx = x^2 +xy - y^2`
Solve the following differential equation.
`(x + a) dy/dx = – y + a`
Solve:
(x + y) dy = a2 dx
Choose the correct alternative:
Differential equation of the function c + 4yx = 0 is
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.
Find the particular solution of the following differential equation
`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.
Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x
∴ `1/"e"^(2y) "d"y` = cos x dx
Integrating, we get
`int square "d"y` = cos x dx
∴ `("e"^(-2y))/(-2)` = sin x + c1
∴ e–2y = – 2sin x – 2c1
∴ `square` = c, where c = – 2c1
This is general solution.
When x = `pi/6`, y = 0, we have
`"e"^0 + 2sin pi/6` = c
∴ c = `square`
∴ particular solution is `square`
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
The value of `dy/dx` if y = |x – 1| + |x – 4| at x = 3 is ______.
