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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

Choose the correct alternative: Differential equation of the function c + 4yx = 0 is - Mathematics and Statistics

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प्रश्न

Choose the correct alternative:

Differential equation of the function c + 4yx = 0 is

पर्याय

  • `xy + ("d"y)/("d"x)` = 0

  • `x ("d"y)/("d"x) + y` = 0

  • `("d"y)/("d"x) - 4xy` =0

  • `x ("d"y)/("d"x) + 1` = 0

MCQ
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उत्तर

`x ("d"y)/("d"x) + y` = 0

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  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 1.8: Differential Equation and Applications - Q.1

संबंधित प्रश्‍न

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


Differential equation \[\frac{dy}{dx} = y, y\left( 0 \right) = 1\]
Function y = ex


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


\[\frac{dy}{dx} = \frac{1 - \cos 2y}{1 + \cos 2y}\]

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

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

x cos2 y  dx = y cos2 x dy


(y + xy) dx + (x − xy2) dy = 0


(y2 + 1) dx − (x2 + 1) dy = 0


Solve the differential equation \[x\frac{dy}{dx} + \cot y = 0\] given that \[y = \frac{\pi}{4}\], when \[x=\sqrt{2}\]


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

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

Solve the following initial value problem:-

\[y' + y = e^x , y\left( 0 \right) = \frac{1}{2}\]


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


Solve the following initial value problem:-

\[\left( 1 + y^2 \right) dx + \left( x - e^{- \tan^{- 1} y} \right) dx = 0, y\left( 0 \right) = 0\]


If the marginal cost of manufacturing a certain item is given by C' (x) = \[\frac{dC}{dx}\] = 2 + 0.15 x. Find the total cost function C (x), given that C (0) = 100.

 

Show that the equation of the curve whose slope at any point is equal to y + 2x and which passes through the origin is y + 2 (x + 1) = 2e2x.


The tangent at any point (x, y) of a curve makes an angle tan−1(2x + 3y) with x-axis. Find the equation of the curve if it passes through (1, 2).


Find the equation to the curve satisfying x (x + 1) \[\frac{dy}{dx} - y\]  = x (x + 1) and passing through (1, 0).


The slope of a curve at each of its points is equal to the square of the abscissa of the point. Find the particular curve through the point (−1, 1).


Find the equation of the curve that passes through the point (0, a) and is such that at any point (x, y) on it, the product of its slope and the ordinate is equal to the abscissa.


The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.


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

`y=sqrt(a^2-x^2)`              `x+y(dy/dx)=0`


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


Solve the following differential equation.

xdx + 2y dx = 0


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


Solve the differential equation `("d"y)/("d"x) + y` = e−x 


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


The integrating factor of the differential equation `"dy"/"dx" (x log x) + y` = 2logx is ______.


An appropriate substitution to solve the differential equation `"dx"/"dy" = (x^2 log(x/y) - x^2)/(xy log(x/y))` is ______.


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