Advertisements
Advertisements
प्रश्न
Solve `x^2 "dy"/"dx" - xy = 1 + cos(y/x)`, x ≠ 0 and x = 1, y = `pi/2`
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove that:
`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`
(1 + x) (1 + y2) dx + (1 + y) (1 + x2) dy = 0
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
x2 dy + y (x + y) dx = 0
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following initial value problem:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]
The rate of growth of a population is proportional to the number present. If the population of a city doubled in the past 25 years, and the present population is 100000, when will the city have a population of 500000?
A bank pays interest by continuous compounding, that is, by treating the interest rate as the instantaneous rate of change of principal. Suppose in an account interest accrues at 8% per year, compounded continuously. Calculate the percentage increase in such an account over one year.
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?
At every point on a curve the slope is the sum of the abscissa and the product of the ordinate and the abscissa, and the curve passes through (0, 1). Find the equation of the curve.
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.
Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.
The differential equation \[x\frac{dy}{dx} - y = x^2\], has the general solution
y2 dx + (x2 − xy + y2) dy = 0
Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
Solve the differential equation:
dr = a r dθ − θ dr
Solve:
(x + y) dy = a2 dx
y2 dx + (xy + x2)dy = 0
`xy dy/dx = x^2 + 2y^2`
The function y = ex is solution ______ of differential equation
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`
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`
Solve the differential equation `"dy"/"dx" + 2xy` = y
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?
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
`d/(dx)(tan^-1 (sqrt(1 + x^2) - 1)/x)` is equal to:
Solve the differential equation
`x + y dy/dx` = x2 + y2
