Advertisements
Advertisements
प्रश्न
Find the general solution of the following differential equation:
`x (dy)/(dx) = y - xsin(y/x)`
उत्तर
We have the differential equation:
`(dy)/(dx) = y/x - sin(y/x)`
The equation is a homogeneous differential equation.
Putting `y = vx ⇒ (dy)/(dx) = v + x (dv)/(dx)`
The differential equation becomes
`v + x (dv)/(dx) = v - sinv`
⇒ `(dv)/(sinv) = - (dx)/x` ⇒ cosecvdv = `-(dx)/x`
Integrating both sides, we get
`log|"cosec"v - cotv| = - log|x| + logK, K > 0` (Here, log K is an arbitrary constant.)
⇒ `log|("cosec"v - cotv)x| = log K`
⇒ `|("cosec"v - cotv)x| = K`
⇒ `("cosec"v - cotv)x = +- K`
⇒ `("cosec" y/x - cot y/x)x = C`, which is the required general solution.
APPEARS IN
संबंधित प्रश्न
Write the integrating factor of the following differential equation:
(1+y2) dx−(tan−1 y−x) dy=0
Find the differential equation of the family of lines passing through the origin.
Find the differential equation representing the family of curves v=A/r+ B, where A and B are arbitrary constants.
Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y2 = a (b2 – x2)
Form a differential equation representing the given family of curves by eliminating arbitrary constants a and b.
y = ex (a cos x + b sin x)
Find the particular solution of the differential equation (1 + e2x) dy + (1 + y2) ex dx = 0, given that y = 1 when x = 0.
Solve the differential equation `ye^(x/y) dx = (xe^(x/y) + y^2)dy, (y != 0)`
Find the differential equation of all the circles which pass through the origin and whose centres lie on x-axis.
The equation of the curve satisfying the differential equation y (x + y3) dx = x (y3 − x) dy and passing through the point (1, 1) is
Verify that xy = a ex + b e−x + x2 is a solution of the differential equation \[x\frac{d^2 y}{d x^2} + 2\frac{dy}{dx} - xy + x^2 - 2 = 0.\]
Show that y = C x + 2C2 is a solution of the differential equation \[2 \left( \frac{dy}{dx} \right)^2 + x\frac{dy}{dx} - y = 0.\]
Show that y2 − x2 − xy = a is a solution of the differential equation \[\left( x - 2y \right)\frac{dy}{dx} + 2x + y = 0.\]
\[\frac{dy}{dx} = y^2 + 2y + 2\]
\[\frac{dy}{dx} = x^2 e^x\]
\[\frac{dy}{dx} - x \sin^2 x = \frac{1}{x \log x}\]
tan y dx + tan x dy = 0
(1 + x) y dx + (1 + y) x dy = 0
cos y log (sec x + tan x) dx = cos x log (sec y + tan y) dy
cosec x (log y) dy + x2y dx = 0
Find the general solution of the differential equation `"dy"/"dx" = y/x`.
The general solution of the differential equation `(dy)/(dx) + x/y` = 0 is
Solution of the equation 3 tan(θ – 15) = tan(θ + 15) is
The general solution of the differential equation `(dy)/(dx) = e^(x + y)` is
The general solution of the differential equation `(ydx - xdy)/y` = 0
Find the general solution of differential equation `(dy)/(dx) = (1 - cosx)/(1 + cosx)`