Advertisements
Advertisements
प्रश्न
Solve the following differential equation.
`(x + a) dy/dx = – y + a`
Advertisements
उत्तर
`(x + a) dy/dx = – y + a`
∴ `dy/dx + y/((x+a)) = a / ((x+a))`
The given equation is of the form
`dy/ dx + py = Q`
where, `P = 1/((x+a)) and Q = a/((x+a))`
∴ I.F. = `e ^(int^(pdx) = e ^(int^(1/(x+a))^dx)`
= `e^(log^ |x+a|) = (x+a)`
∴ Solution of the given equation is
`y ( I.F.) = int Q (I.F.) dx + c `
∴ `y(x + a) = int a/((x+a)) (x+a) dx + c`
∴ `y(x + a) = a int 1 dx + c`
∴ y (x + a) = ax + 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.
Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
(1 − x2) dy + xy dx = xy2 dx
y (1 + ex) dy = (y + 1) ex dx
Solve the following differential equation:
\[\text{ cosec }x \log y \frac{dy}{dx} + x^2 y^2 = 0\]
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
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 decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.
Find the equation of the curve which passes through the point (2, 2) and satisfies the differential equation
\[y - x\frac{dy}{dx} = y^2 + \frac{dy}{dx}\]
The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y \sin x = 1\], is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.
The differential equation `y dy/dx + x = 0` represents family of ______.
For the following differential equation find the particular solution.
`(x + 1) dy/dx − 1 = 2e^(−y)`,
when y = 0, x = 1
Solve the following differential equation.
`dy/dx + y` = 3
Solve the differential equation xdx + 2ydy = 0
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
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`
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
If `y = log_2 log_2(x)` then `(dy)/(dx)` =
