हिंदी

For the differential equation given, find a particular solution satisfying the given condition: dydx-3ycotx=sin2x;y=2 when x=π2 - Mathematics

Advertisements
Advertisements

प्रश्न

For the differential equation given, find a particular solution satisfying the given condition:

`dy/dx - 3ycotx = sin 2x; y = 2`  when `x = pi/2`

योग
Advertisements

उत्तर

The given equation is 

`dy/dx - 3 y cot x = sin 2x`                ....(1)

Which is a linear equation of the type

`dy/dx + Py = Q`

Here P = - 3cot x and Q =  sin 2x

∴ `intP dx = -3 int cot x  dx = -3 log |sin x|`

∴ `I.F. = e^(-3log|sin x|)`

`= e^(log cosec^3 x)`

`= cosec^3 x`

∴ The solution is `y. (I.F.) = int Q. (I.F.)  dx + C`

`y cosec^3 x = int sin2x cosec^3x dx + C`

`= int (2 sin x cos x)/(sin^3 x)  dx + C`

`= 2 int cosec x cot x  dx + C`

`= - 2 cosec  x  +C`

⇒ y = -2 sin2 x + C sin3 x                          ....(2)

When `x = pi/2, y = 2`

∴ `2 = -2 sin^2  pi/2 + C sin^3  pi/2`

⇒ 2 = -2 (1)2 + C (1)3

⇒ C = 2 + 2

⇒ C = 4

Putting in (2), we get

y = - 2sin2 x + 4 sin3 x

⇒ y = 4 sin3 x - 2 sin x

Which is the required solution.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Exercise 9.6 [पृष्ठ ४१४]

APPEARS IN

एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 9 Differential Equations
Exercise 9.6 | Q 15 | पृष्ठ ४१४

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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 + 3y^2) dy/dx = y(y > 0)`


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.


Solve the differential equation `(tan^(-1) x- y) dx = (1 + x^2) dy`


Find the general solution of the differential equation `dy/dx - y = sin x`


x dy = (2y + 2x4 + x2) dx


\[\left( x^2 - 1 \right)\frac{dy}{dx} + 2\left( x + 2 \right)y = 2\left( x + 1 \right)\]

\[x\frac{dy}{dx} + 2y = x \cos x\]

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:-
\[\left( 1 + x^2 \right)\frac{dy}{dx} - 2xy = \left( x^2 + 2 \right)\left( x^2 + 1 \right)\]


Find the integerating factor of the differential equation `xdy/dx - 2y = 2x^2` . 


Solve the following differential equation:

`cos^2 "x" * "dy"/"dx" + "y" = tan "x"`


Solve the following differential equation:

`"x" "dy"/"dx" + "2y" = "x"^2 * log "x"`


Solve the following differential equation:

y dx + (x - y2) dy = 0


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"`.


Form the differential equation of all circles which pass through the origin and whose centers lie on X-axis.


`(x + 2y^3 ) dy/dx = y`


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 sin y `("dy"/"dx")` = cos y(1 - x cos y) is ______.


Integrating factor of `dy/dx + y = x^2 + 5` is ______ 


The equation x2 + yx2 + x + y = 0 represents


The integrating factor of differential equation `(1 - y)^2  (dx)/(dy) + yx = ay(-1 < y < 1)`


Let y = y(x), x > 1, be the solution of the differential equation `(x - 1)(dy)/(dx) + 2xy = 1/(x - 1)`, with y(2) = `(1 + e^4)/(2e^4)`. If y(3) = `(e^α + 1)/(βe^α)`, then the value of α + β is equal to ______.


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 ______.


Let y = f(x) be a real-valued differentiable function on R (the set of all real numbers) such that f(1) = 1. If f(x) satisfies xf'(x) = x2 + f(x) – 2, then the area bounded by f(x) with x-axis between ordinates x = 0 and x = 3 is equal to ______.


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)`


If sec x + tan x is the integrating factor of `dy/dx + Py` = Q, then value of P is ______.


The slope of the tangent to the curve x = sin θ and y = cos 2θ at θ = `π/6` is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×