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Find the order and degree of the following differential equation: ddddddd2ydx2+y+(dydx-d3ydx3)32 = 0 - Business Mathematics and Statistics

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प्रश्न

Find the order and degree of the following differential equation:

`("d"^2y)/("d"x^2) + y + (("d"y)/("d"x) - ("d"^3y)/("d"x^3))^(3/2)` = 0

योग

उत्तर

`("d"^2y)/("d"x^2) + y = - (("d"y)/("d"x) - ("d"^3y)/("d"x^3))^(3/2)`

Squaring on both sides

`(("d"^2y)/("d"x^2) + y)^2 = {-(("d"y)/("d"x) - ("d"^3y)/("d"x^3))^(3/2)}`

`(("d"^2y)/("d"x^2) + y)^2 = (("d"y)/("d"x) - ("d"^3y)/("d"x^3))^3`

Highest order derivative is `("d"^3y)/("d"x^3)`

∴ Order = 3

Power of the highest order derivative `("d"^3y)/("d"x^3)` is 3

∴ Degree = 3

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Formation of Ordinary Differential Equations
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Differential Equations - Exercise 4.1 [पृष्ठ ८५]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 4 Differential Equations
Exercise 4.1 | Q 1. (v) | पृष्ठ ८५
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