Advertisements
Advertisements
प्रश्न
Select and write the correct alternative from the given option for the question
The differential equation of y = Ae5x + Be–5x is
पर्याय
`("d"^2y)/("d"x^2)` = 25y
`("d"^2y)/("d"x^2)` = – 25y
`("d"^2y)/("d"x^2)` = 5y
`y ("d"^2y)/("d"x^2)` = – 5y
Advertisements
उत्तर
`("d"^2y)/("d"x^2)` = 25y
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`
Verify that y = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[y = \left( \frac{dy}{dx} \right)^2\]
|
\[y = \frac{1}{4} \left( x \pm a \right)^2\]
|
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
(sin x + cos x) dy + (cos x − sin x) dx = 0
(1 + x2) dy = xy dx
If y(x) is a solution of the different equation \[\left( \frac{2 + \sin x}{1 + y} \right)\frac{dy}{dx} = - \cos x\] and y(0) = 1, then find the value of y(π/2).
y ex/y dx = (xex/y + y) dy
(y2 − 2xy) dx = (x2 − 2xy) dy
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \left( x + 1 \right) e^{- x} , y\left( 1 \right) = 0\]
Solve the following initial value problem:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
Find the equation of the curve which passes through the origin and has the slope x + 3y− 1 at any point (x, y) on it.
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.
Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.
Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
The differential equation satisfied by ax2 + by2 = 1 is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
Verify that the function y = e−3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + \frac{dy}{dx} - 6y = 0.\]
Solve the differential equation:
`"x"("dy")/("dx")+"y"=3"x"^2-2`
In the following example, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| xy = log y + k | y' (1 - xy) = y2 |
Determine the order and degree of the following differential equations.
| Solution | D.E. |
| y = 1 − logx | `x^2(d^2y)/dx^2 = 1` |
Solve the following differential equation.
xdx + 2y dx = 0
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
Solve the differential equation:
`e^(dy/dx) = x`
For the differential equation, find the particular solution (x – y2x) dx – (y + x2y) dy = 0 when x = 2, y = 0
Solve the following differential equation y log y = `(log y - x) ("d"y)/("d"x)`
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
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
Solve: ydx – xdy = x2ydx.
