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प्रश्न
Show that y = a cos bx is a solution of the! differential equation `("d"^2y)/("d"x^2) + "b"^2y` = 0
उत्तर
Given y = a cos bx .......(1)
Differentiating equation (1) w.r.t ‘x’, we get
`("d"y)/("d"x)` = a(– sin bx)b
`("d"y)/("d"x)` = – ab sin bx
Again differentiating, we get
`("d"^2y)/("d"x^2) = - "ab" cos "b"x * "b"`
y = a cos bx
`y/"a"` = cos bx
`("d"^2y)/("d"x^2) = - "ab"^2 cos "b"x`
`("d"^2y)/("d"x^2) = - "ab"^2 xx y/"a"`
`("d"^2y)/("d"x^2) = - "b"^2y`
`("d"^2y)/("d"x^2)` = 0
Therefore, y = a cos bx is a solution of the differential equation `("d"^2y)/("d"x^2) + "b"^2y` = 0
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