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प्रश्न
If y = eax. cos bx, then prove that
`(d^2y)/(dx^2) - 2ady/dx + (a^2 + b^2)y` = 0
उत्तर
y = eax. cos bx
`dy/dx = ae^(ax).cosbx - be^(ax).sinbx ...(i)`
`dy/dx = ay - be^(ax).sinbx`
`(d^2y)/(dx^2) = ady/dx - b(ae^(ax).sinbx + be^(ax).cosbx)`
`(d^2y)/(dx^2) = ady/dx - abe^(ax).sinbx - b^2e^(ax).cosbx`
`(d^2y)/(dx^2) = ady/dx - a(ay - dy/dx) - b^2y ` ...[Substituting beax sin bx from (i)]
`(d^2y)/(dx^2) = ady/dx - a^2y + ady/dx - b^2y`
`therefore (d^2y)/(dx^2) - 2ady/dx + (a^2 + b^2)y` = 0
Hence Proved
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