Let `y=f(x)` satisfies the equation
`f(x) = (e^(-x)+e^(x))cosx-2x+int_(0)^(x)(x-t)f^(')(t)dt`
The value of `f(0)+f^(')(0)` equal (a) -1 (b) 0 (c) 1 (d) none of these

Options

A. `-1`
B. 0
C. 1
D. 1

Answer

B

Explanation

For problem 13 to 15
We have `f(x) = (e^(-x)+e^(x))cosx-2x-int_(0)^(x) (x-t)f^(')(t)dt`
`rArr f(x) = (e^(x)+e^(-x)) cosx-2x-[x int_(0)^(x)f^(')(t)dt-int_(0)^(x)tf^(')(t)dt]`
`rArr f(x) = (e^(x)+e^(-x))cosx-2x -[(xf(x)-f(0))-{[t.f(t)]_(0)^(x)-int_(0)^(x)f(t)dt}]`
`rArr f(x) = (e^(x) + e^(-x)) cos-int_(0)^(x)f(t)dt` ..............(i)
Differentiating w.r.t x,
`rArr f^(')(x)+f(x)=cosx(e^(x)-e^(-x))-(e^(x)+e^(-x))sinx`.............(ii)
or `(dy)/(dx) + y=e^(x)(cosx-sinx) - e^(-x)(cosx+sinx)`..............(iii)
Above is linear differential equation
I.F. `=e^(x)`
`therefore` Solution is
`y.e^(x) = inte^(2x)(cosx-sinx)dx-int(cosx+sinx)dx`
`rArr y.e^(x)= inte^(2x)(cosx-sinx)dx-(sinx-cosx)+C`
`rArr y.e^(x) = e^(2x) (3/5cosx-1/5sinx) -(sinx-cosx)+C`
When `x=0, y=2`
`rArr 2=3/5+1+C`
`rArr =2/5`
`therefore y=e^(x) (3/5 cosx-1/5sinx) - (sinx-cosx)e^(-x)+2/5e^(-x)`