हिंदी

Integrating factor of the differential equation dydxxdydx-y = sinx is ______.

Advertisements
Advertisements

प्रश्न

Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is ______.

रिक्त स्थान भरें
Advertisements

उत्तर

Integrating factor of the differential equation `x "dy"/"dx" - y` = sinx is `1/x`.

Explanation:

`1/x`; given differential equation can be written as `"dy"/"dx" - y/x = sinx/x`

And Therefore I.F. = `"e"^(int 1/x "d"x)`

= `"e"^(-logx)`

= `1/x`.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Differential Equations - Solved Examples [पृष्ठ १८९]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 9 Differential Equations
Solved Examples | Q 22. (vi) | पृष्ठ १८९

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

\[\frac{d^2 y}{d x^2} + \left( \frac{dy}{dx} \right)^2 + xy = 0\]

Verify that y = \[\frac{a}{x} + b\] is a solution of the differential equation
\[\frac{d^2 y}{d x^2} + \frac{2}{x}\left( \frac{dy}{dx} \right) = 0\]


Verify that y2 = 4ax is a solution of the differential equation y = x \[\frac{dy}{dx} + a\frac{dx}{dy}\]


Verify that \[y = ce^{tan^{- 1}} x\]  is a solution of the differential equation \[\left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + \left( 2x - 1 \right)\frac{dy}{dx} = 0\]


For the following differential equation verify that the accompanying function is a solution:

Differential equation Function
\[x\frac{dy}{dx} + y = y^2\]
\[y = \frac{a}{x + a}\]

Differential equation \[\frac{d^2 y}{d x^2} + y = 0, y \left( 0 \right) = 0, y' \left( 0 \right) = 1\] Function y = sin x


\[\frac{dy}{dx} = \tan^{- 1} x\]


(sin x + cos x) dy + (cos x − sin x) dx = 0


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

Solve the differential equation \[\frac{dy}{dx} = e^{x + y} + x^2 e^y\].

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

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

\[2x\frac{dy}{dx} = 3y, y\left( 1 \right) = 2\]

\[\frac{dr}{dt} = - rt, r\left( 0 \right) = r_0\]

\[2\left( y + 3 \right) - xy\frac{dy}{dx} = 0\], y(1) = −2

Find the particular solution of the differential equation \[\frac{dy}{dx} = - 4x y^2\]  given that y = 1, when x = 0.


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

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

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

\[\left[ x\sqrt{x^2 + y^2} - y^2 \right] dx + xy\ dy = 0\]

Find the particular solution of the differential equation \[\frac{dy}{dx} = \frac{xy}{x^2 + y^2}\] given that y = 1 when x = 0.

 


Solve the following initial value problem:-
\[x\frac{dy}{dx} - y = \log x, y\left( 1 \right) = 0\]


In a culture, the bacteria count is 100000. The number is increased by 10% in 2 hours. In how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present?


The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when


The differential equation satisfied by ax2 + by2 = 1 is


What is integrating factor of \[\frac{dy}{dx}\] + y sec x = tan x?


In the following verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:-

y = ex + 1            y'' − y' = 0


Form the differential equation representing the family of parabolas having vertex at origin and axis along positive direction of x-axis.


If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`


For each of the following differential equations find the particular solution.

`y (1 + logx)dx/dy - x log x = 0`,

when x=e, y = e2.


y2 dx + (xy + x2)dy = 0


 `dy/dx = log x`


Solve the differential equation sec2y tan x dy + sec2x tan y dx = 0


Solve the following differential equation `("d"y)/("d"x)` = x2y + y


Solve the following differential equation

`yx ("d"y)/("d"x)` = x2 + 2y2 


For the differential equation, find the particular solution

`("d"y)/("d"x)` = (4x +y + 1), when y = 1, x = 0


Solve the following differential equation `("d"y)/("d"x)` = cos(x + y)

Solution: `("d"y)/("d"x)` = cos(x + y)    ......(1)

Put `square`

∴ `1 + ("d"y)/("d"x) = "dv"/("d"x)`

∴ `("d"y)/("d"x) = "dv"/("d"x) - 1`

∴ (1) becomes `"dv"/("d"x) - 1` = cos v

∴ `"dv"/("d"x)` = 1 + cos v

∴ `square` dv = dx

Integrating, we get

`int 1/(1 + cos "v")  "d"v = int  "d"x`

∴ `int 1/(2cos^2 ("v"/2))  "dv" = int  "d"x`

∴ `1/2 int square  "dv" = int  "d"x`

∴ `1/2* (tan("v"/2))/(1/2)` = x + c

∴ `square` = x + c


Solve the differential equation `"dy"/"dx"` = 1 + x + y2 + xy2, when y = 0, x = 0.


`d/(dx)(tan^-1  (sqrt(1 + x^2) - 1)/x)` is equal to:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×