Advertisements
Advertisements
प्रश्न
Determine the order and degree of the following differential equations.
| Solution | D.E |
| y = aex + be−x | `(d^2y)/dx^2= 1` |
Advertisements
उत्तर
y = aex + be−x ......(1)
Differentiating w.r.t. x, we get
`dy/dx = ae^x - be^-x`
Again, differentiating w.r.t. x, we get
`(d^2y)/dx^2 = ae^x - be^-x`
∴ `(d^2y)/dx^2 = y` .....[From (i)]
∴ Given function is a solution of the given differential equation.
APPEARS IN
संबंधित प्रश्न
Show that Ax2 + By2 = 1 is a solution of the differential equation x \[\left\{ y\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 \right\} = y\frac{dy}{dx}\]
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
Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]
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).
y ex/y dx = (xex/y + y) dy
Solve the following initial value problem:-
\[\frac{dy}{dx} + 2y = e^{- 2x} \sin x, y\left( 0 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} - 3y \cot x = \sin 2x; y = 2\text{ when }x = \frac{\pi}{2}\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).
Write the differential equation obtained by eliminating the arbitrary constant C in the equation x2 − y2 = C2.
Find the solution of the differential equation
\[x\sqrt{1 + y^2}dx + y\sqrt{1 + x^2}dy = 0\]
The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is
Solve the following differential equation.
xdx + 2y dx = 0
Choose the correct alternative.
Bacteria increases at the rate proportional to the number present. If the original number M doubles in 3 hours, then the number of bacteria will be 4M in
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.
`xy dy/dx = x^2 + 2y^2`
Solve the differential equation xdx + 2ydy = 0
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.
Given that `"dy"/"dx" = "e"^-2x` and y = 0 when x = 5. Find the value of x when y = 3.
The differential equation (1 + y2)x dx – (1 + x2)y dy = 0 represents a family of:
Solve the differential equation
`y (dy)/(dx) + x` = 0
