Advertisements
Advertisements
Question
Integrating factor of the differential equation `(1 - x^2) ("d"y)/("d"x) - xy` = 1 is ______.
Options
– x
`x/(1 + x^2)`
`sqrt(1 - x^2)`
`1/2 log (1 - x^2)`
Solution
Integrating factor of the differential equation `(1 - x^2) ("d"y)/("d"x) - xy` = 1 is `sqrt(1 - x^2)`.
Explanation:
The given differential equation is `(1 - x^2) ("d"y)/("d"x) - xy` = 1
⇒ `("d"y)/("d"x) - x/(1 - x^2) * y = 1/(1 - x^2)`
Here, P = `x/(1 - x^2)` and Q = `1/(1 - x^2)`
∴ Integrating factor I.F. = `"e"^(int Pdx)`
= `"e"^(int (-x)/(1 - x^2) dx)`
= `"e"^(1/2 log(1 - x^2))`
= `sqrt(1 - x^2)`
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 + y/x = x^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 dy/dx + y - x + xy cot x = 0(x != 0)`
\[\frac{dy}{dx}\] + y cos x = sin x cos x
Solve the differential equation \[\left( x + 2 y^2 \right)\frac{dy}{dx} = y\], given that when x = 2, y = 1.
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 differential equation \[\frac{dy}{dx}\] + y cot x = 2 cos x, given that y = 0 when x = \[\frac{\pi}{2}\] .
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:
`("x + a")"dy"/"dx" - 3"y" = ("x + a")^5`
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.
Find the equation of the curve passing through the point `(3/sqrt2, sqrt2)` having a slope of the tangent to the curve at any point (x, y) is -`"4x"/"9y"`.
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
Form the differential equation of all circles which pass through the origin and whose centers lie on X-axis.
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 equation x2 + yx2 + x + y = 0 represents
If y = y(x) is the solution of the differential equation, `(dy)/(dx) + 2ytanx = sinx, y(π/3)` = 0, then the maximum value of the function y (x) over R is equal to ______.
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 the solution curve y = y(x) of the differential equation y2dx + (x2 – xy + y2)dy = 0, which passes through the point (1, 1) and intersects the line y = `sqrt(3) x` at the point `(α, sqrt(3) α)`, then value of `log_e (sqrt(3)α)` is equal to ______.