Advertisements
Advertisements
Question
Solve the following differential equation.
`(dθ)/dt = − k (θ − θ_0)`
Advertisements
Solution
`(dθ)/dt = − k (θ − θ_0)`, k is constant.
∴`(dθ)/ (θ − θ_0) = -k dt`
Integrating on both sides, we get
`int(dθ)/ (θ − θ_0) = -k int dt`
∴ log |θ − θ0| = - kt + c
∴ θ − θ0 = `e ^(-kt+c)`
APPEARS IN
RELATED QUESTIONS
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x + y\frac{dy}{dx} = 0\]
|
\[y = \pm \sqrt{a^2 - x^2}\]
|
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
Solve the following differential equations:
\[\frac{dy}{dx} = \frac{y}{x}\left\{ \log y - \log x + 1 \right\}\]
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 tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).
The normal to a given curve at each point (x, y) on the curve passes through the point (3, 0). If the curve contains the point (3, 4), find its equation.
The differential equation satisfied by ax2 + by2 = 1 is
The integrating factor of the differential equation \[x\frac{dy}{dx} - y = 2 x^2\]
y2 dx + (x2 − xy + y2) dy = 0
If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`
Solve the following differential equation.
`xy dy/dx = x^2 + 2y^2`
Solve the differential equation:
`e^(dy/dx) = x`
y2 dx + (xy + x2)dy = 0
For the differential equation, find the particular solution
`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0
A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution
State whether the following statement is True or False:
The integrating factor of the differential equation `("d"y)/("d"x) - y` = x is e–x
Solve the following differential equation
sec2 x tan y dx + sec2 y tan x dy = 0
Solution: sec2 x tan y dx + sec2 y tan x dy = 0
∴ `(sec^2x)/tanx "d"x + square` = 0
Integrating, we get
`square + int (sec^2y)/tany "d"y` = log c
Each of these integral is of the type
`int ("f'"(x))/("f"(x)) "d"x` = log |f(x)| + log c
∴ the general solution is
`square + log |tan y|` = log c
∴ log |tan x . tan y| = log c
`square`
This is the general solution.
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
Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.
Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.
