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A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution. - Mathematics and Statistics

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प्रश्न

A solution of a differential equation which can be obtained from the general solution by giving particular values to the arbitrary constants is called ___________ solution.

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उत्तर

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

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differential Equation and Applications - Miscellaneous Exercise 8 [पृष्ठ १७२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 2.3 | पृष्ठ १७२

संबंधित प्रश्न

Show that the function y = A cos 2x − B sin 2x is a solution of the differential equation \[\frac{d^2 y}{d x^2} + 4y = 0\].


Verify that y = cx + 2c2 is a solution of the differential equation 

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

\[\frac{dy}{dx} = \frac{1 + y^2}{y^3}\]

(1 + x2) dy = xy dx


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

(y + xy) dx + (x − xy2) dy = 0


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

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

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

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

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

(x + y) (dx − dy) = dx + dy


Solve the following initial value problem:
\[\frac{dy}{dx} + y \cot x = 4x\text{ cosec }x, y\left( \frac{\pi}{2} \right) = 0\]


A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is \[y^2 - 2xy\frac{dy}{dx} - x^2 = 0\], and hence find the curve.


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


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

Solution D.E.
xy = log y + k y' (1 - xy) = y2

Solve the following differential equation.

`dy/dx = x^2 y + y`


The solution of `dy/dx + x^2/y^2 = 0` is ______


Solve the differential equation:

`e^(dy/dx) = x`


x2y dx – (x3 + y3) dy = 0


`xy dy/dx  = x^2 + 2y^2`


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

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


Solve `("d"y)/("d"x) = (x + y + 1)/(x + y - 1)` when x = `2/3`, y = `1/3`


Solve the following differential equation

`x^2  ("d"y)/("d"x)` = x2 + xy − y2 


The function y = cx is the solution of differential equation `("d"y)/("d"x) = y/x`


Verify y = log x + c is the solution of differential equation `x ("d"^2y)/("d"x^2) + ("d"y)/("d"x)` = 0


Solve the following differential equation 

sec2 x tan y dx + sec2 y tan x dy = 0

Solution: sec2 x tan y dx + sec2 y tan x dy = 0

∴ `(sec^2x)/tanx  "d"x + square` = 0

Integrating, we get

`square + int (sec^2y)/tany  "d"y` = log c

Each of these integral is of the type

`int ("f'"(x))/("f"(x))  "d"x` = log |f(x)| + log c

∴ the general solution is

`square + log |tan y|` = log c

∴ log |tan x . tan y| = log c

`square`

This is the general solution.


Solution of `x("d"y)/("d"x) = y + x tan  y/x` is `sin(y/x)` = cx


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