Advertisements
Advertisements
प्रश्न
For each of the following differential equations find the particular solution.
(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0
उत्तर
(x − y2 x)dx − (y + x2 y) dy = 0, when x = 2, y = 0
∴ x(1- y2) dx = y(1 + x2 ) dy
∴ `(xdx)/(1+x^2) = (ydy)/(1-y^2)`
Integrating on both sides, we get
`int( 2x)/(1+x^2) dx = int(2y)/(1-y^2 )dy`
∴ `int( 2x)/(1+x^2) dx = - int(-2y)/(1-y^2 )dy`
∴ `log | 1 + x^2| = -log| 1-y^2| + log |c|`
∴ `log |1 + x^2 | = log |c /(1-y^2)|`
∴ (1 + x 2) ( 1 - y2 ) = c …(i)
When x = 2, y = 0, we have
(1 + 4) (1 - 0) = c
∴ c = 5
Substituting c = 5 in (i),we get
(1 + x2) ( 1-y2 ) = 5,
which is the required particular solution.
APPEARS IN
संबंधित प्रश्न
(1 − x2) dy + xy dx = xy2 dx
tan y dx + sec2 y tan x dy = 0
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
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?
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.
y2 dx + (x2 − xy + y2) dy = 0
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
Solve: `("d"y)/("d"x) + 2/xy` = x2
Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.
