हिंदी

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

Advertisements
Advertisements

प्रश्न

Solve the following differential equation.

y dx + (x - y2 ) dy = 0

योग
Advertisements

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]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differential Equation and Applications - Exercise 8.5 [पृष्ठ १६८]

APPEARS IN

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

संबंधित प्रश्न

\[\frac{d^3 x}{d t^3} + \frac{d^2 x}{d t^2} + \left( \frac{dx}{dt} \right)^2 = e^t\]

\[\frac{d^2 y}{d x^2} + 4y = 0\]

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

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


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


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


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

Differential equation Function
\[x^3 \frac{d^2 y}{d x^2} = 1\]
\[y = ax + b + \frac{1}{2x}\]

\[\frac{1}{x}\frac{dy}{dx} = \tan^{- 1} x, x \neq 0\]

\[\frac{dy}{dx} = \cos^3 x \sin^2 x + x\sqrt{2x + 1}\]

\[\sin\left( \frac{dy}{dx} \right) = k ; y\left( 0 \right) = 1\]

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

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

y ex/y dx = (xex/y + y) dy


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

 

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.


Find the equation of the curve passing through the point (0, 1) if the slope of the tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and the ordinate of the point.


Define a differential equation.


The solution of the differential equation \[\frac{dy}{dx} = \frac{ax + g}{by + f}\] represents a circle when


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


Solve the following differential equation : \[\left( \sqrt{1 + x^2 + y^2 + x^2 y^2} \right) dx + xy \ dy = 0\].


If a + ib = `("x" + "iy")/("x" - "iy"),` prove that `"a"^2 +"b"^2 = 1` and `"b"/"a" = (2"xy")/("x"^2 - "y"^2)`


Determine the order and degree of the following differential equations.

Solution D.E
y = aex + be−x `(d^2y)/dx^2= 1`

Form the differential equation from the relation x2 + 4y2 = 4b2


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


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


Solve the following differential equation y log y = `(log  y - x) ("d"y)/("d"x)`


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


The differential equation of all non horizontal lines in a plane is `("d"^2x)/("d"y^2)` = 0


Solve the differential equation `dy/dx + xy = xy^2` and find the particular solution when y = 4, x = 1.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×