Advertisements
Advertisements
प्रश्न
The solution of `(1 + x^2) ("d"y)/("d"x) + 2xy - 4x^2` = 0 is ______.
Advertisements
उत्तर
The solution of `(1 + x^2) ("d"y)/("d"x) + 2xy - 4x^2` = 0 is y = `4/3 x^3/((1 + x^2)) + "c" (1 + x^2)^-1`.
Explanation:
The given differential equation is `(1 + x^2) ("d"y)/("d"x) + 2xy - 4x^2` = 0
⇒ `("d"y)/("d"x) + (2xy)/(1 + x^2) = (4x^2)/(1 + x^2)`
Since it is a linear differential equation
∴ P = `(2x)/(1 + x^2)` and Q = `(4x^2)/(1 + x^2)`
Integrating factor I.F. = `"e"^(int Pdx)`
= `"e"^(int (2x)/(1 + x^2) "d"x)`
= `"e"^(log(1 + x^2))`
= `(1 + x^2)`
∴ Solution is `y xx "I"."F" = int "Q" xx "I"."F". "d"x + "c"`
⇒ `y xx (1 + x^2) = int (4x)/(1 + x^2) xx (1 + x^2)"d"x + "c"`
⇒ `y xx (1 + x^2) = int 4x^2 "d"x + "c"`
⇒ `y xx (1 + x^2) = 4/3 x^3 + "c"`
⇒ y = `4/3 x^3/((1 + x^2)) + "c"(1 + x^2)^-1`
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:
`x log x dy/dx + y= 2/x log x`
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.
The Integrating Factor of the differential equation `dy/dx - y = 2x^2` is ______.
Solve the differential equation `x dy/dx + y = x cos x + sin x`, given that y = 1 when `x = pi/2`
x dy = (2y + 2x4 + x2) dx
(x + tan y) dy = sin 2y dx
Find the general solution of the differential equation \[\frac{dy}{dx} - y = \cos x\]
Solve the differential equation \[\left( y + 3 x^2 \right)\frac{dx}{dy} = x\]
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 + x^2) dy/dx + 2xy - 4x^2 = 0,` subject to the initial condition y(0) = 0.
Solve the following differential equation:
`cos^2 "x" * "dy"/"dx" + "y" = tan "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.
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.
The integrating factor of the differential equation sin y `("dy"/"dx")` = cos y(1 - x cos y) is ______.
The integrating factor of the differential equation (1 + x2)dt = (tan-1 x - t)dx is ______.
Integrating factor of the differential equation `(1 - x^2) ("d"y)/("d"x) - xy` = 1 is ______.
The equation x2 + yx2 + x + y = 0 represents
The integrating factor of the differential equation `x (dy)/(dx) - y = 2x^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 ______.
If the slope of the tangent at (x, y) to a curve passing through `(1, π/4)` is given by `y/x - cos^2(y/x)`, then the equation of the curve is ______.
Find the general solution of the differential equation:
`(x^2 + 1) dy/dx + 2xy = sqrt(x^2 + 4)`
