Advertisements
Advertisements
प्रश्न
Solve the following differential equation `("d"y)/("d"x)` = x2y + y
Advertisements
उत्तर
`("d"y)/("d"x)` = x2y + y
= (x2 + 1)y
∴ `1/y "d"y` = (x2 + 1) dx
Integrating on both sides, we get
`int 1/y "d"y = int(x^2 + 1) "d"x`
∴ log |y| = `x^3/3 + x + c`
APPEARS IN
संबंधित प्रश्न
Assume that a rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.
(y2 + 1) dx − (x2 + 1) dy = 0
Solve the following differential equation:
\[y e^\frac{x}{y} dx = \left( x e^\frac{x}{y} + y^2 \right)dy, y \neq 0\]
2xy dx + (x2 + 2y2) dy = 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:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
In a simple circuit of resistance R, self inductance L and voltage E, the current `i` at any time `t` is given by L \[\frac{di}{dt}\]+ R i = E. If E is constant and initially no current passes through the circuit, prove that \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
Find the equation of the curve passing through the point \[\left( 1, \frac{\pi}{4} \right)\] and tangent at any point of which makes an angle tan−1 \[\left( \frac{y}{x} - \cos^2 \frac{y}{x} \right)\] with x-axis.
Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\] = x (x + 1) and passing through (1, 0).
The solution of the differential equation y1 y3 = y22 is
Which of the following is the integrating factor of (x log x) \[\frac{dy}{dx} + y\] = 2 log x?
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
Choose the correct option from the given alternatives:
The solution of `1/"x" * "dy"/"dx" = tan^-1 "x"` is
In each of the following examples, verify that the given function is a solution of the corresponding differential equation.
| Solution | D.E. |
| y = ex | `dy/ dx= y` |
For each of the following differential equations find the particular solution.
(x − y2 x) dx − (y + x2 y) dy = 0, when x = 2, y = 0
Solve the following differential equation.
`(x + y) dy/dx = 1`
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
The solution of `dy/ dx` = 1 is ______.
Solve the differential equation:
dr = a r dθ − θ dr
`xy dy/dx = x^2 + 2y^2`
The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`
Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0
y = `a + b/x`
`(dy)/(dx) = square`
`(d^2y)/(dx^2) = square`
Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`
= `x square + 2 square`
= `square`
Hence y = `a + b/x` is solution of `square`
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`
