Let y(x) be a function satisfying `d^(2)y//dx^(2)-dy//dx+e^(2x)=0,y(0)=2 and y'(0)=1.` If maximum value of y(x) is `y(alpha)`, Then Integral part of `(2alpha)` is ……

Answer

1

Explanation

We have `(d^(2)y)/(dx^(2))-(dy)/(dx)+e^(2x)=0`
Put `(dy)/(dx)=t`
`therefore (dt)/(dx) -t=-e^(2x)`, which is linear differential equation
`rArr` solution is `te^(-x)=-inte^(2x)e^(-x)dx+c`
`rArr (dy)/(dx) e^(-x)=-e^(x)+c`
`rArr y^(')(0)=1 rArr c=2`
`therefore y=2e^(x)-e^(2x)/2+c^(')`
`therefore y(x) = 2e^(x)- e^(2x)/2+1/2`
`y(x) le5/2` for `x=log_(2)2`
So, `[2alpha]=[log_(e)4]=1`