Advertisements
Advertisements
प्रश्न
The solution of differential equation coty dx = xdy is ______.
उत्तर
The solution of differential equation coty dx = xdy is x = C sec y.
Explanation:
The given differential equation is cot y dx = x dy
⇒ `("d"y)/(cot y) = ("d"x)/x`
⇒ tan y dy = `("d"x)/x`
Integrating both sides, we get
`int tan y "d"y = int ("d"x)/x`
⇒ `log sec y = log x + log "c"`
⇒ `log sec y - log x = log "c"`
⇒ `log|(sec y)/x| = log "C"`
∴ `secy/x` = C
⇒ `x/(sec y) = 1/"C"`
⇒ `x/secy` = C ....`[1/"c" = "C"]`
∴ x = C sec y
APPEARS IN
संबंधित प्रश्न
The differential equation of `y=c/x+c^2` is :
(a)`x^4(dy/dx)^2-xdy/dx=y`
(b)`(d^2y)/dx^2+xdy/dx+y=0`
(c)`x^3(dy/dx)^2+xdy/dx=y`
(d)`(d^2y)/dx^2+dy/dx-y=0`
The solution of the differential equation dy/dx = sec x – y tan x is:
(A) y sec x = tan x + c
(B) y sec x + tan x = c
(C) sec x = y tan x + c
(D) sec x + y tan x = c
The differential equation of the family of curves y=c1ex+c2e-x is......
(a)`(d^2y)/dx^2+y=0`
(b)`(d^2y)/dx^2-y=0`
(c)`(d^2y)/dx^2+1=0`
(d)`(d^2y)/dx^2-1=0`
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = x2 + 2x + C : y′ – 2x – 2 = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
y = cos x + C : y′ + sin x = 0
Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:
`y sqrt(1 + x^2) : y' = (xy)/(1+x^2)`
Solve the differential equation `[e^(-2sqrtx)/sqrtx - y/sqrtx] dx/dy = 1 (x != 0).`
if `y = sin^(-1) (6xsqrt(1-9x^2))`, `1/(3sqrt2) < x < 1/(3sqrt2)` then find `(dy)/(dx)`
The solution of the differential equation x dx + y dy = x2 y dy − y2 x dx, is
The solution of the differential equation \[\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2}\], is
The number of arbitrary constants in the particular solution of a differential equation of third order is
Find the particular solution of the differential equation `(1+y^2)+(x-e^(tan-1 )y)dy/dx=` given that y = 0 when x = 1.
x (e2y − 1) dy + (x2 − 1) ey dx = 0
`(2ax+x^2)(dy)/(dx)=a^2+2ax`
(x3 − 2y3) dx + 3x2 y dy = 0
\[y^2 + \left( x + \frac{1}{y} \right)\frac{dy}{dx} = 0\]
Solve the following differential equation:-
\[x\frac{dy}{dx} + 2y = x^2 , x \neq 0\]
Solve the following differential equation:-
\[x\frac{dy}{dx} + 2y = x^2 \log x\]
Find the equation of a curve passing through the point (0, 0) and whose differential equation is \[\frac{dy}{dx} = e^x \sin x.\]
Solve the differential equation: `(d"y")/(d"x") - (2"x")/(1+"x"^2) "y" = "x"^2 + 2`
Solve the differential equation: ` ("x" + 1) (d"y")/(d"x") = 2e^-"y" - 1; y(0) = 0.`
Solve: `2(y + 3) - xy "dy"/"dx"` = 0, given that y(1) = – 2.
Solution of the differential equation tany sec2xdx + tanx sec2ydy = 0 is ______.
The solution of `x ("d"y)/("d"x) + y` = ex is ______.
The solution of the equation (2y – 1)dx – (2x + 3)dy = 0 is ______.
The solution of `("d"y)/("d"x) + y = "e"^-x`, y(0) = 0 is ______.
The solution of `("d"y)/("d"x) = (y/x)^(1/3)` is `y^(2/3) - x^(2/3)` = c.
Find the general solution of the differential equation `x (dy)/(dx) = y(logy - logx + 1)`.
If the solution curve of the differential equation `(dy)/(dx) = (x + y - 2)/(x - y)` passes through the point (2, 1) and (k + 1, 2), k > 0, then ______.