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प्रश्न
Show that the function y = ax + 2a2 is a solution of the differential equation `2(dy/dx)^2 + x(dy/dx) - y = 0`.
बेरीज
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उत्तर
y = ax + 2a2
Differentiate w.r.t. x
`dy/dx = a(1) = 0`
`dy/dx = a`
`2(dy/dx)^2 + x(dy/dx) - y = 2(a)^2 + x(a) - y`
= 2a2 + ax – (ax + 2a2) [y = ax + 2a2]
= 2a2 + ax – ax – 2a2
= 0
`2(dy/dx)^2 + x(dy/dx) - y = 0`
Hence, y = ax + 2a2 is a solution of the differential equation `2(dy/dx)^2 + x(dy/dx) - y = 0`.
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