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
Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
(ey + 1) cos x dx + ey sin x dy = 0
(y + xy) dx + (x − xy2) 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 = \[\frac{\pi}{2}\]
x2 dy + y (x + y) dx = 0
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
The surface area of a balloon being inflated, changes at a rate proportional to time t. If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
The equation of the curve whose slope is given by \[\frac{dy}{dx} = \frac{2y}{x}; x > 0, y > 0\] and which passes through the point (1, 1) is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
Solve the following differential equation : \[y^2 dx + \left( x^2 - xy + y^2 \right)dy = 0\] .
The price of six different commodities for years 2009 and year 2011 are as follows:
| Commodities | A | B | C | D | E | F |
|
Price in 2009 (₹) |
35 | 80 | 25 | 30 | 80 | x |
| Price in 2011 (₹) | 50 | y | 45 | 70 | 120 | 105 |
The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.
The differential equation `y dy/dx + x = 0` represents family of ______.
Form the differential equation from the relation x2 + 4y2 = 4b2
Solve the following differential equation.
(x2 − y2 ) dx + 2xy dy = 0
The integrating factor of the differential equation `dy/dx - y = x` is e−x.
`xy dy/dx = x^2 + 2y^2`
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Solve the following differential equation
`yx ("d"y)/("d"x)` = x2 + 2y2
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
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`
Given that `"dy"/"dx"` = yex and x = 0, y = e. Find the value of y when x = 1.
Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
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)` =
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
