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प्रश्न
A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.
उत्तर
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called particular solution.
संबंधित प्रश्न
Show that y = ax3 + bx2 + c is a solution of the differential equation \[\frac{d^3 y}{d x^3} = 6a\].
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x\frac{dy}{dx} + y = y^2\]
|
\[y = \frac{a}{x + a}\]
|
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
(1 − x2) dy + xy dx = xy2 dx
In a bank principal increases at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years (e0.5 = 1.648).
In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
The differential equation satisfied by ax2 + by2 = 1 is
Form the differential equation representing the family of curves y = a sin (x + b), where a, b are arbitrary constant.
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.
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Choose the correct alternative.
The solution of `x dy/dx = y` log y is
Solve the differential equation:
dr = a r dθ − θ dr
`xy dy/dx = x^2 + 2y^2`
Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`
Solve the differential equation (x2 – yx2)dy + (y2 + xy2)dx = 0
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
