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प्रश्न
The differential equation of the family of curves y=c1ex+c2e-x is......
(a)`(d^2y)/dx^2+y=0`
(b)`(d^2y)/dx^2-y=0`
(c)`(d^2y)/dx^2+1=0`
(d)`(d^2y)/dx^2-1=0`
उत्तर
y=c1ex+c2e-x
differentiate w.r.t 'x'
`dy/dx=c_1e^x-c_2e^(-x) ..............(1)`
differentiate equation (1) w.r.t 'x'
`(d^2y)/(dx^2)=c_1e^x+c_2e^(-x)`
`(d^2y)/(dx^2)-y=0`
=L.H.S
`=c_1e^x+c_2e^(-x)-(c_1e^x+c_2e^(-x))`
`=c_1e^x+c_2e^(-x)-c_1e^x-c_2e^(-x)`
`=0`
Hence
`(b) (d^2y)/(dx^2)-y=0`
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