Advertisements
Advertisements
प्रश्न
The solution of `dy/dx + x^2/y^2 = 0` is ______
पर्याय
x3 + y3 = 7
x2 + y2 = c
x3 + y3 = c
x + y = c
Advertisements
उत्तर
The solution of `dy/dx + x^2/y^2 = 0` is x3 + y3 = c
APPEARS IN
संबंधित प्रश्न
Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.
(1 − x2) dy + xy dx = xy2 dx
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
y ex/y dx = (xex/y + y) dy
\[x^2 \frac{dy}{dx} = x^2 + xy + y^2 \]
Solve the following initial value problem:-
\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
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 differential equation whose general solution is
x3 + y3 = 35ax.
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
Choose the correct alternative.
The integrating factor of `dy/dx - y = e^x `is ex, then its solution is
Solve the differential equation xdx + 2ydy = 0
Solve the following differential equation y2dx + (xy + x2) dy = 0
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log y
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
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
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
