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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

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 = 4 sin 3x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 9y = 0\]


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

xy dy = (y − 1) (x + 1) dx


tan y \[\frac{dy}{dx}\] = sin (x + y) + sin (x − y) 

 


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

Find the solution of the differential equation cos y dy + cos x sin y dx = 0 given that y = \[\frac{\pi}{2}\], when x = \[\frac{\pi}{2}\] 

 


Solve the following initial value problem:-

\[\frac{dy}{dx} + y\cot x = 2\cos x, y\left( \frac{\pi}{2} \right) = 0\]


The slope of the tangent at a point P (x, y) on a curve is \[\frac{- x}{y}\]. If the curve passes through the point (3, −4), find the equation of the curve.


Find the equation of the curve such that the portion of the x-axis cut off between the origin and the tangent at a point is twice the abscissa and which passes through the point (1, 2).


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.


Integrating factor of the differential equation cos \[x\frac{dy}{dx} + y\] sin x = 1, is


The solution of the differential equation \[\frac{dy}{dx} - \frac{y\left( x + 1 \right)}{x} = 0\] is given by


Find the coordinates of the centre, foci and equation of directrix of the hyperbola x2 – 3y2 – 4x = 8.


Solve the following differential equation.

xdx + 2y dx = 0


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


The differential equation of `y = k_1e^x+ k_2 e^-x` is ______.


Verify y = `a + b/x` is solution of `x(d^2y)/(dx^2) + 2 (dy)/(dx)` = 0

y = `a + b/x`

`(dy)/(dx) = square`

`(d^2y)/(dx^2) = square`

Consider `x(d^2y)/(dx^2) + 2(dy)/(dx)`

= `x square + 2 square`

= `square`

Hence y = `a + b/x` is solution of `square`


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


A man is moving away from a tower 41.6 m high at a rate of 2 m/s. If the eye level of the man is 1.6 m above the ground, then the rate at which the angle of elevation of the top of the tower changes, when he is at a distance of 30 m from the foot of the tower, is


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