Advertisements
Advertisements
Question
For the differential equation, find the general solution:
`x dy/dx + y - x + xy cot x = 0(x != 0)`
Advertisements
Solution
Given differential equation
`x dy/dx + y - x + xy cot x = 0`
⇒ `x dy/dx + y (1 + x cot x) = x`
or `dy/dx + (1/x + cot x) y = 1` ...(i)
Comparing with `dy/dx + Py = Q`
`P = 1/x + cot x` and Q = 1
∴ `I.F. = e^(int P dx) = e^(int(1/x + cot x)dx)`
`= e^(log x) + log sin x`
`=> e^(log (x sin x)) = x sin x`
Hence the required solution
∴ `y × I.F. = int I.F. xx Q dx + C`
`=> y xx x sin x = int 1 * x sin x dx + C`
`=> xy sin x = - x cos x + int 1 cos x dx + C`
`=> xy sin x = - x cos x + sin x + C`
⇒ y = `(- x cos x)/(x sin x) + (sin x)/(x sin x) + C/(x sin x)`
⇒ `y = 1/x - cot x + C/ (x sin x)`
APPEARS IN
RELATED QUESTIONS
Find the the differential equation for all the straight lines, which are at a unit distance from the origin.
For the differential equation, find the general solution:
`dy/dx + 2y = sin x`
For the differential equation, find the general solution:
`dy/dx + 3y = e^(-2x)`
For the differential equation, find the general solution:
`dy/dx + y/x = x^2`
For the differential equation, find the general solution:
`(x + y) dy/dx = 1`
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:
`(1 + x^2)dy/dx + 2xy = 1/(1 + x^2); y = 0` when x = 1
x dy = (2y + 2x4 + x2) dx
dx + xdy = e−y sec2 y dy
\[\frac{dy}{dx}\] = y tan x − 2 sin x
\[\frac{dy}{dx}\] + y cos x = sin x cos x
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`
Find the integerating factor of the differential equation `xdy/dx - 2y = 2x^2` .
If f(x) = x + 1, find `"d"/"dx"("fof") ("x")`
Solve the following differential equation:
`cos^2 "x" * "dy"/"dx" + "y" = tan "x"`
Solve the following differential equation:
`"dy"/"dx" + "y" * sec "x" = tan "x"`
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:
dr + (2r cotθ + sin2θ)dθ = 0
Solve the following differential equation:
y dx + (x - y2) dy = 0
The integrating factor of `(dy)/(dx) + y` = e–x is ______.
`(x + 2y^3 ) dy/dx = y`
The slope of the tangent to the curves x = 4t3 + 5, y = t2 - 3 at t = 1 is ______
Which of the following is a second order differential equation?
The integrating factor of the differential equation `x (dy)/(dx) - y = 2x^2` is
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 ______.
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 ______.
The slope of the tangent to the curve x = sin θ and y = cos 2θ at θ = `π/6` is ______.
