Advertisements
Advertisements
प्रश्न
Solve the following differential equation.
y dx + (x - y2 ) dy = 0
Advertisements
उत्तर
y dx + (x - y2 ) dy = 0
∴ y dx = (y2 - x) dy
∴ `dx/dy = (y^2 - x) /y= y - x/y `
∴ `dx/dy + x/y = y`
The given equation is of the form
`dx/dy + Px = Q`
where, P = `1/y` and Q = y
∴ I.F. = `e int^ (pdy) = e int ^(1/ydy) = e ^(log |y|)= y`
∴ Solution of the given equation is
`x (I.F.) =int Q (I.F.) dy + c_1`
∴ `xy = int y(y) dy = y^3/3 + c_1`
∴ 3xy = y3 + c …[3c1 = c]
APPEARS IN
संबंधित प्रश्न
Find the differential equation of all the parabolas with latus rectum '4a' and whose axes are parallel to x-axis.
Show that the function y = A cos x + B sin x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + y = 0\]
Show that y = ex (A cos x + B sin x) is the solution of the differential equation \[\frac{d^2 y}{d x^2} - 2\frac{dy}{dx} + 2y = 0\]
Verify that y2 = 4a (x + a) is a solution of the differential equations
\[y\left\{ 1 - \left( \frac{dy}{dx} \right)^2 \right\} = 2x\frac{dy}{dx}\]
Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x
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
x cos y dy = (xex log x + ex) dx
dy + (x + 1) (y + 1) dx = 0
Solve the following differential equation:
\[y\left( 1 - x^2 \right)\frac{dy}{dx} = x\left( 1 + y^2 \right)\]
x2 dy + y (x + y) dx = 0
2xy dx + (x2 + 2y2) dy = 0
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 \[i = \frac{E}{R}\left\{ 1 - e^{- \left( R/L \right)t} \right\}.\]
Show that all curves for which the slope at any point (x, y) on it is \[\frac{x^2 + y^2}{2xy}\] are rectangular hyperbola.
Which of the following differential equations has y = C1 ex + C2 e−x as the general solution?
Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].
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
y dx – x dy + log x dx = 0
Solve the following differential equation
`x^2 ("d"y)/("d"x)` = x2 + xy − y2
Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)
Solution: `("d"y)/("d"x)` = cos(x + y) ......(1)
Put `square`
∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`
∴ (1) becomes `"dv"/("d"x) - 1` = cos v
∴ `"dv"/("d"x)` = 1 + cos v
∴ `square` dv = dx
Integrating, we get
`int 1/(1 + cos "v") "d"v = int "d"x`
∴ `int 1/(2cos^2 ("v"/2)) "dv" = int "d"x`
∴ `1/2 int square "dv" = int "d"x`
∴ `1/2* (tan("v"/2))/(1/2)` = x + c
∴ `square` = x + c
