Advertisements
Advertisements
प्रश्न
In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-
y = ex + 1 y'' − y' = 0
Advertisements
उत्तर
We have,
y'' − y' = 0 ............(1)
Now,
y = ex +1
⇒ y'= ex
⇒ y'' = ex
Putting the above values in (1), we get
LHS = ex − ex = 0 = RHS
Thus, y = ex + 1 is the solution of the given differential equation.
APPEARS IN
संबंधित प्रश्न
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 \[\frac{d^2 y}{d x^2} + y = 0\]
Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\] satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]
For the following differential equation verify that the accompanying function is a solution:
| Differential equation | Function |
|
\[x^3 \frac{d^2 y}{d x^2} = 1\]
|
\[y = ax + b + \frac{1}{2x}\]
|
Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex
dy + (x + 1) (y + 1) dx = 0
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 differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.
\[\frac{dy}{dx} = \frac{y}{x} + \sin\left( \frac{y}{x} \right)\]
Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]
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:-
\[\frac{dy}{dx} + y \tan x = 2x + x^2 \tan x, y\left( 0 \right) = 1\]
The rate of increase of bacteria in a culture is proportional to the number of bacteria present and it is found that the number doubles in 6 hours. Prove that the bacteria becomes 8 times at the end of 18 hours.
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.
If sin x is an integrating factor of the differential equation \[\frac{dy}{dx} + Py = Q\], then write the value of P.
The integrating factor of the differential equation (x log x)
\[\frac{dy}{dx} + y = 2 \log x\], is given by
The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when
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 following differential equation.
`(x + a) dy/dx = – y + a`
State whether the following is True or False:
The degree of a differential equation is the power of the highest ordered derivative when all the derivatives are made free from negative and/or fractional indices if any.
Solve: `("d"y)/("d"x) + 2/xy` = x2
Solve the following differential equation
`y log y ("d"x)/("d"y) + x` = log 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.
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`
An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.
Solve the differential equation
`x + y dy/dx` = x2 + y2
