Advertisements
Advertisements
प्रश्न
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
विकल्प
y" + y' = 0
y" − ω2 y = 0
y" = −ω2 y
y" + y = 0
Advertisements
उत्तर
y" = −ω2 y
We have,
y = A cos ωt + B sin ωt .....(1)
Differentiating both sides of (1) with respect to x, we get
\[\frac{dy}{dt} = - A\omega \sin \omega t + B \omega \cos \omega t\] .....(2)
Differentiating both sides of (2) again with respect to x, we get
\[\frac{d^2 y}{d t^2} = - A \omega^2 \cos \omega t - B \omega^2 \sin \omega t\]
\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 \left( A \cos \omega t + B \sin \omega t \right)\]
\[ \Rightarrow \frac{d^2 y}{d t^2} = - \omega^2 y ..........\left[ \text{Using }\left( 1 \right) \right]\]
\[ \therefore y'' = - \omega^2 y\]
APPEARS IN
संबंधित प्रश्न
If 1, `omega` and `omega^2` are the cube roots of unity, prove `(a + b omega + c omega^2)/(c + s omega + b omega^2) = omega^2`
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
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 \[\frac{dy}{dx} + y = 2, y \left( 0 \right) = 3\] Function y = e−x + 2
(sin x + cos x) dy + (cos x − sin x) dx = 0
C' (x) = 2 + 0.15 x ; C(0) = 100
x cos y dy = (xex log x + ex) dx
(1 − x2) dy + xy dx = xy2 dx
(y2 − 2xy) dx = (x2 − 2xy) dy
Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]
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.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Write the differential equation obtained eliminating the arbitrary constant C in the equation xy = C2.
Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is
The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Solve the following differential equation.
`y^3 - dy/dx = x dy/dx`
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 + 2xy = x`
`dy/dx = log x`
Select and write the correct alternative from the given option for the question
Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
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.
Solution of `x("d"y)/("d"x) = y + x tan y/x` is `sin(y/x)` = cx
The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
