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Solve the following differential equation. y dx + (x - y 2 ) dy = 0 - Mathematics and Statistics

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प्रश्न

Solve the following differential equation.

y dx + (x - y2 ) dy = 0

बेरीज

उत्तर

y dx + (x - y2 ) dy = 0

∴ y dx = (y2 - x) dy

∴ `dx/dy = (y^2 - x) /y=  y - x/y `

∴ `dx/dy + x/y = y`

The given equation is of the form

`dx/dy + Px = Q`

where, P = `1/y` and Q = y

∴ I.F. = `e int^ (pdy) = e int ^(1/ydy) = e ^(log |y|)= y`

∴ Solution of the given equation is

`x (I.F.) =int Q (I.F.) dy + c_1`

∴  `xy = int y(y) dy = y^3/3 + c_1`

∴  3xy = y3 + c   …[3c1 = c]

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पाठ 8: Differential Equation and Applications - Exercise 8.5 [पृष्ठ १६८]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] 12 Standard HSC Maharashtra State Board
पाठ 8 Differential Equation and Applications
Exercise 8.5 | Q 1.5 | पृष्ठ १६८

संबंधित प्रश्‍न

\[\sqrt{1 + \left( \frac{dy}{dx} \right)^2} = \left( c\frac{d^2 y}{d x^2} \right)^{1/3}\]

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

Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.


Verify that y = log \[\left( x + \sqrt{x^2 + a^2} \right)^2\]  satisfies the differential equation \[\left( a^2 + x^2 \right)\frac{d^2 y}{d x^2} + x\frac{dy}{dx} = 0\]


\[\frac{dy}{dx} + \frac{1 + y^2}{y} = 0\]

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

 


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

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

Solve the following initial value problem:-

\[dy = \cos x\left( 2 - y\text{ cosec }x \right)dx\]


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`

Choose the correct alternative.

The integrating factor of `dy/dx -  y = e^x `is ex, then its solution is


State whether the following is True or False:

The integrating factor of the differential equation `dy/dx - y = x` is e-x


x2y dx – (x3 + y3) dy = 0


`xy dy/dx  = x^2 + 2y^2`


 `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 `("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


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


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


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