हिंदी

Find the order and degree of the following differential equation: [d3ydx3+x]32=d2ydx2

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प्रश्न

Find the order and degree of the following differential equation:

`[ (d^3y)/dx^3 + x]^(3/2) = (d^2y)/dx^2`

योग
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उत्तर

`[ d^3y/dx^3 + x]^(3/2) = (d^2y)/dx^2`

Squaring on both sides, we get

`[(d^3y)/dx^3 + x]^3 = ((d^2y)/dx^2)^2`

By definition of order and degree,

Order : 3 ; Degree : 3

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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 4.01 | पृष्ठ १७२

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