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

A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution

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प्रश्न

A solution of differential equation which can be obtained from the general solution by giving particular values to the arbitrary constant is called ______ solution

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उत्तर

particular

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पाठ 1.8: Differential Equation and Applications - Q.2

संबंधित प्रश्‍न

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

Show that the differential equation of which \[y = 2\left( x^2 - 1 \right) + c e^{- x^2}\]  is a solution is \[\frac{dy}{dx} + 2xy = 4 x^3\]


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

Differential equation Function
\[y = \left( \frac{dy}{dx} \right)^2\]
\[y = \frac{1}{4} \left( x \pm a \right)^2\]

\[x\left( x^2 - 1 \right)\frac{dy}{dx} = 1, y\left( 2 \right) = 0\]

x cos2 y  dx = y cos2 x dy


\[\frac{dy}{dx} + \frac{\cos x \sin y}{\cos y} = 0\]

Solve the differential equation \[\frac{dy}{dx} = \frac{2x\left( \log x + 1 \right)}{\sin y + y \cos y}\], given that y = 0, when x = 1.


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


\[\frac{dy}{dx} + 1 = e^{x + y}\]

x2 dy + y (x + y) dx = 0


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

 

Solve the following initial value problem:-

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


Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]


The decay rate of radium at any time t is proportional to its mass at that time. Find the time when the mass will be halved of its initial mass.


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).


The x-intercept of the tangent line to a curve is equal to the ordinate of the point of contact. Find the particular curve through the point (1, 1).


Write the differential equation representing the family of straight lines y = Cx + 5, where C is an arbitrary constant.


The differential equation obtained on eliminating A and B from y = A cos ωt + B sin ωt, is


The price of six different commodities for years 2009 and year 2011 are as follows: 

Commodities A B C D E F

Price in 2009 (₹)

35 80 25 30 80 x
Price in 2011 (₹) 50 y 45 70 120 105

The Index number for the year 2011 taking 2009 as the base year for the above data was calculated to be 125. Find the values of x andy if the total price in 2009 is ₹ 360.


In the following example, verify that the given function is a solution of the corresponding differential equation.

Solution D.E.
y = xn `x^2(d^2y)/dx^2 - n xx (xdy)/dx + ny =0`

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.


Solve the following differential equation.

y2 dx + (xy + x2 ) dy = 0


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


y2 dx + (xy + x2)dy = 0


Select and write the correct alternative from the given option for the question

Bacterial increases at the rate proportional to the number present. If original number M doubles in 3 hours, then number of bacteria will be 4M in


The function y = ex is solution  ______ of differential equation


Find the particular solution of the following differential equation

`("d"y)/("d"x)` = e2y cos x, when x = `pi/6`, y = 0.

Solution: The given D.E. is `("d"y)/("d"x)` = e2y cos x

∴ `1/"e"^(2y)  "d"y` = cos x dx

Integrating, we get

`int square  "d"y` = cos x dx

∴ `("e"^(-2y))/(-2)` = sin x + c1

∴ e–2y = – 2sin x – 2c1

∴ `square` = c, where c = – 2c

This is general solution.

When x = `pi/6`, y = 0, we have

`"e"^0 + 2sin  pi/6` = c

∴ c = `square`

∴ particular solution is `square`


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


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