Advertisements
Advertisements
प्रश्न
Form the differential equation of all circles which pass through the origin and whose centers lie on X-axis.
Advertisements
उत्तर
Let C(h, 0) be the center of the circle which passes through the origin. Then the radius of the circle is h.
∴ equation of the circle is (x - h)2 + (y - 0)2 = h2
∴ x2 - 2hx + h + y2 = h2
∴ x2 + h2 = 2hx ....(1)
Differentiating both sides w.r.t. x, we get
`"2x" + "2y" "dy"/"dx" = "2h"`
Substituting the value of 2h in equation (1), we get
`"x"^2 + "y"^2 = ("2x" + "2y" "dy"/"dx")"x"`
∴ `"x"^2 + "y"^2 = 2"x"^2 + 2"xy" "dy"/"dx"`
∴ `"2xy" "dy"/"dx" + "x"^2 - "y"^2 = 0`
This is the required D.E.
APPEARS IN
संबंधित प्रश्न
For the differential equation, find the general solution:
`dy/dx + 3y = e^(-2x)`
For the differential equation, find the general solution:
`dy/dx + (sec x) y = tan x (0 <= x < pi/2)`
For the differential equation, find the general solution:
`x dy/dx + 2y= x^2 log x`
For the differential equation, find the general solution:
`x log x dy/dx + y= 2/x log x`
For the differential equation, find the general solution:
(1 + x2) dy + 2xy dx = cot x dx (x ≠ 0)
For the differential equation, find the general solution:
y dx + (x – y2) dy = 0
For the differential equation given, find a particular solution satisfying the given condition:
`dy/dx - 3ycotx = sin 2x; y = 2` when `x = pi/2`
Find the equation of the curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangent to the curve at that point by 5.
The integrating factor of the differential equation.
`(1 - y^2) dx/dy + yx = ay(-1 < y < 1)` is ______.
\[\frac{dy}{dx}\] + y cos x = sin x cos x
Find the particular solution of the differential equation \[\frac{dx}{dy} + x \cot y = 2y + y^2 \cot y, y ≠ 0\] given that x = 0 when \[y = \frac{\pi}{2}\].
Solve the following differential equation:- \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\]
Solve the following differential equation: \[\left( \cot^{- 1} y + x \right) dy = \left( 1 + y^2 \right) dx\] .
Find the integerating factor of the differential equation `x(dy)/(dx) - 2y = 2x^2`
Solve the differential equation: (1 +x2 ) dy + 2xy dx = cot x dx
Solve the differential equation: `(1 + x^2) dy/dx + 2xy - 4x^2 = 0,` subject to the initial condition y(0) = 0.
Solve the following differential equation:
`dy/dx + y/x = x^3 - 3`
Solve the following differential equation:
`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`
Solve the following differential equation:
`("x + y") "dy"/"dx" = 1`
Solve the following differential equation:
`("x + a")"dy"/"dx" - 3"y" = ("x + a")^5`
Solve the following differential equation:
dr + (2r cotθ + sin2θ)dθ = 0
Solve the following differential equation:
`(1 + "x"^2) "dy"/"dx" + "y" = "e"^(tan^-1 "x")`
Find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it.
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
Find the general solution of the equation `("d"y)/("d"x) - y` = 2x.
Solution: The equation `("d"y)/("d"x) - y` = 2x
is of the form `("d"y)/("d"x) + "P"y` = Q
where P = `square` and Q = `square`
∴ I.F. = `"e"^(int-"d"x)` = e–x
∴ the solution of the linear differential equation is
ye–x = `int 2x*"e"^-x "d"x + "c"`
∴ ye–x = `2int x*"e"^-x "d"x + "c"`
= `2{x int"e"^-x "d"x - int square "d"x* "d"/("d"x) square"d"x} + "c"`
= `2{x ("e"^-x)/(-1) - int ("e"^-x)/(-1)*1"d"x} + "c"`
∴ ye–x = `-2x*"e"^-x + 2int"e"^-x "d"x + "c"`
∴ e–xy = `-2x*"e"^-x+ 2 square + "c"`
∴ `y + square + square` = cex is the required general solution of the given differential equation
The integrating factor of the differential equation (1 + x2)dt = (tan-1 x - t)dx is ______.
The slope of the tangent to the curves x = 4t3 + 5, y = t2 - 3 at t = 1 is ______
The solution of `(1 + x^2) ("d"y)/("d"x) + 2xy - 4x^2` = 0 is ______.
The integrating factor of the differential equation `x (dy)/(dx) - y = 2x^2` is
The integrating factor of differential equation `(1 - y)^2 (dx)/(dy) + yx = ay(-1 < y < 1)`
State whether the following statement is true or false.
The integrating factor of the differential equation `(dy)/(dx) + y/x` = x3 is – x.
Let y = y(x) be a solution curve of the differential equation (y + 1)tan2xdx + tanxdy + ydx = 0, `x∈(0, π/2)`. If `lim_(x→0^+)` xy(x) = 1, then the value of `y(π/2)` is ______.
Let y = y(x) be the solution curve of the differential equation `(dy)/(dx) + ((2x^2 + 11x + 13)/(x^3 + 6x^2 + 11x + 6)) y = ((x + 3))/(x + 1), x > - 1`, which passes through the point (0, 1). Then y(1) is equal to ______.
Let the solution curve y = y(x) of the differential equation (4 + x2) dy – 2x (x2 + 3y + 4) dx = 0 pass through the origin. Then y (2) is equal to ______.
If sin x is the integrating factor (IF) of the linear differential equation `dy/dx + Py` = Q then P is ______.
Solve the differential equation `dy/dx+2xy=x` by completing the following activity.
Solution: `dy/dx+2xy=x` ...(1)
This is the linear differential equation of the form `dy/dx +Py =Q,"where"`
`P=square` and Q = x
∴ `I.F. = e^(intPdx)=square`
The solution of (1) is given by
`y.(I.F.)=intQ(I.F.)dx+c=intsquare dx+c`
∴ `ye^(x^2) = square`
This is the general solution.
The slope of tangent at any point on the curve is 3. lf the curve passes through (1, 1), then the equation of curve is ______.
Solve:
`xsinx dy/dx + (xcosx + sinx)y` = sin x
