Advertisements
Advertisements
प्रश्न
Solve the following differential equation:
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
Advertisements
उत्तर
\[\left( 1 + y^2 \right) \tan^{- 1} xdx + 2y\left( 1 + x^2 \right)dy = 0\]
\[ \Rightarrow \left( 1 + y^2 \right) \tan^{- 1} xdx = - 2y\left( 1 + x^2 \right)dy\]
\[ \Rightarrow \frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \int\frac{\tan^{- 1} x}{\left( 1 + x^2 \right)}dx = - \int\frac{2y}{\left( 1 + y^2 \right)}dy\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} = - \log\left| 1 + y^2 \right| + C\]
\[ \Rightarrow \frac{\left( \tan^{- 1} x \right)^2}{2} + \log\left| 1 + y^2 \right| = C\]
APPEARS IN
संबंधित प्रश्न
Prove that:
`int_0^(2a)f(x)dx = int_0^af(x)dx + int_0^af(2a - x)dx`
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 y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]
Verify that \[y = ce^{tan^{- 1}} x\] is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]
Differential equation \[\frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + 2y = 0, y \left( 0 \right) = 1, y' \left( 0 \right) = 3\] Function y = ex + e2x
(sin x + cos x) dy + (cos x − sin x) dx = 0
x cos y dy = (xex log x + ex) dx
Solve the differential equation \[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\], given that y = 1, when x = 0.
Find the particular solution of the differential equation
(1 – y2) (1 + log x) dx + 2xy dy = 0, given that y = 0 when x = 1.
x2 dy + y (x + y) dx = 0
(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:-
\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]
Solve the following initial value problem:-
\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]
Solve the following initial value problem:-
\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]
The rate of increase in the number of bacteria in a certain bacteria culture is proportional to the number present. Given the number triples in 5 hrs, find how many bacteria will be present after 10 hours. Also find the time necessary for the number of bacteria to be 10 times the number of initial present.
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.
The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by
If xmyn = (x + y)m+n, prove that \[\frac{dy}{dx} = \frac{y}{x} .\]
The price of six different commodities for years 2009 and year 2011 are as follows:
| Commodities | A | B | C | D | E | F |
|
Price in 2009 (₹) |
35 | 80 | 25 | 30 | 80 | x |
| Price in 2011 (₹) | 50 | y | 45 | 70 | 120 | 105 |
The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
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 |
Form the differential equation from the relation x2 + 4y2 = 4b2
Solve the following differential equation.
y2 dx + (xy + x2 ) dy = 0
Solve the following differential equation.
x2y dx − (x3 + y3) dy = 0
The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.
The solution of `dy/dx + x^2/y^2 = 0` is ______
y2 dx + (xy + x2)dy = 0
The function y = ex is solution ______ of differential equation
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`
