हिंदी

Determine the order and degree of the following differential equations. 1+1(dydx)2=(dydx)32

Advertisements
Advertisements

प्रश्न

Determine the order and degree of the following differential equations.

`sqrt(1+1/(dy/dx)^2) = (dy/dx)^(3/2)`

योग
Advertisements

उत्तर

`sqrt(1+1/(dy/dx)^2) = (dy/dx)^(3/2)`

Squaring on both sides, we get

`1 + 1/(dy/dx)^2 = (dy/dx)^3`

∴ `(dy/dx)^2 +1 = (dy/dx)^5`

By definition of order and degree,

Order : 1 ; Degree : 5

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Differential Equation and Applications - Exercise 8.1 [पृष्ठ १६२]

APPEARS IN

बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
अध्याय 8 Differential Equation and Applications
Exercise 8.1 | Q 1.5 | पृष्ठ १६२

संबंधित प्रश्न

Determine the order and degree (if defined) of the differential equation:

y' + 5y = 0


For the given below, verify that the given function (implicit or explicit) is a solution to the corresponding differential equation.

`x^2 = 2y^2 log y : (x^2  + y^2) dy/dx - xy = 0`


\[s^2 \frac{d^2 t}{d s^2} + st\frac{dt}{ds} = s\]

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

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

Write the degree of the differential equation
\[a^2 \frac{d^2 y}{d x^2} = \left\{ 1 + \left( \frac{dy}{dx} \right)^2 \right\}^{1/4}\]


Write the degree of the differential equation \[x^3 \left( \frac{d^2 y}{d x^2} \right)^2 + x \left( \frac{dy}{dx} \right)^4 = 0\]


The order of the differential equation whose general solution is given by y = c1 cos (2x + c2) − (c3 + c4) ax + c5 + c6 sin (x − c7) is


The order of the differential equation satisfying
\[\sqrt{1 - x^4} + \sqrt{1 - y^4} = a\left( x^2 - y^2 \right)\] is


The order of the differential equation \[2 x^2 \frac{d^2 y}{d x^2} - 3\frac{dy}{dx} + y = 0\], is


Determine the order and degree (if defined) of the following differential equation:-

y"' + 2y" + y' = 0


Determine the order and degree (if defined) of the following differential equation:-

y" + (y')2 + 2y = 0


Determine the order and degree of the following differential equation:

`("d"^2"y")/"dx"^2 + "dy"/"dx" + "x" = sqrt(1 + ("d"^3"y")/"dx"^3)`


Determine the order and degree of the following differential equations.

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


Choose the correct alternative.

The order and degree of `(dy/dx)^3 - (d^3y)/dx^3 + ye^x = 0` are respectively.


State whether the following is True or False:

The degree of the differential equation `e^((dy)/(dx)) = dy/dx +c` is not defined.


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

The order and degree of `(1 + (("d"y)/("d"x))^3)^(2/3) = 8 ("d"^3y)/("d"x^3)` are respectively


State whether the following statement is True or False:

Order and degree of differential equation `x ("d"^3y)/("d"x^3) + 6(("d"^2y)/("d"x^2))^2 + y` = 0 is (2, 2)


The order and degree of the differential equation `[1 + 1/("dy"/"dx")^2]^(5/3) = 5 ("d"^2y)/"dx"^2` are respectively.


The third order differential equation is ______ 


The degree of the differential equation `("d"^2y)/("d"x^2) + 3("dy"/"dx")^2 = x^2 log(("d"^2y)/("d"x^2))` is ______.


Order of the differential equation representing the family of ellipses having centre at origin and foci on x-axis is two.


Degree of the differential equation `sqrt(1 + ("d"^2y)/("d"x^2)) = x + "dy"/"dx"` is not defined.


The degree of the differential equation `sqrt(1 + (("d"y)/("d"x))^2)` = x is ______.


Determine the order and degree of the following differential equation:

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


The order and degree of the differential equation `sqrt((dy)/(dx)) - 4 (dy)/(dx) - 7x = 0` are respectively ______.


For the differential equation \[xy \frac{d^{2}y}{dx^{2}} + x \left( \frac{dy}{dx} \right)^{2} - y \frac{dy}{dx} = 0\], what is the degree?


Based on the differential equation illustrated with \[\left(\frac{d^{3}y}{dx^{3}}\right)^{2} + 5\frac{dy}{dx} = \sin(x)\], what is the order?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×