Let y=y(x) be the solution of the differential equation, `dy/dx+y tan x=2x+x^(2)tanx, x in(-pi/2,pi/2),` such that y(0)= 1. Then (a) `y'(pi/4)-y'(-pi/4)=pi-sqrt 2` (b) `y'(pi/4)+y'(-pi/4)=-sqrt 2` (c) `y(pi/4)+y(-pi/4)=-pi^(2)/2+2.` (d) `y(pi/4)-y(-pi/4)=sqrt 2`

Options

A. (a) `y'(pi/4)-y'(-pi/4)=pi-sqrt 2`
B. (b) `y'(pi/4)+y'(-pi/4)=-sqrt 2`
C. (c) `y(pi/4)+y(-pi/4)=-pi^(2)/2+2.`
D. (d) `y(pi/4)-y(-pi/4)=sqrt 2`

Answer

(a)

Explanation

Given differential equation is
`dy/dx=y tan x=2x+x^(2) tan x,` which is linear differential
equation in the from of `dy/dx+Py =Q.`
Here, `P=tan x and Q=2x+x^(2)tan x `
`therefore IF = e^(int tan x dx)=e^(log_(2)(sec x))=sec x`
Now, solution of linear differenital equation is given as
`yxxIF =int(QxxIF)dx+C`
`therefore y(sec x) =int (2x+x^(2) tan x)sec x dx +C`
`= int (2x sec x) dx+ int x^(2) sec x tan x dx +C`
`therefore int x^(2) sec x tan x dx = x^(2) sec x-int (2xsec x) dx`
Therefore, solution is
`y sec x = 2 int x sec x dx+ x^(2) sec x-2int x sec x dx + C`
`rArr y sec x = x^(2) sec x + C … (i)`

`therefore y(0) =1rArr 1(1) = 0(1) +CrArrC=1`
Now,`y=x^(2)+cos x`
and `y'=2x-sin x`
According to options,
`y'(pi/4)-y'((-pi)/4)=(2(pi/4)-1/sqrt2)`
`-(2(-pi/4)+1/sqrt2)=pi-sqrt2`
and`y'(pi/4)+y'(-pi/4)=(2(pi/4)-1/sqrt2) +(2(-pi/4)+1/sqrt2)=0`
and`y'(pi/4)+y'(-pi/4)=pi^(2) /16+1/sqrt2+pi^(2)/16+1/sqrt2=pi^(2)/4+sqrt2`
and`y(pi/4)-y(-pi/4)=pi^(2) /16+1/sqrt2 -pi^(2)/16 -1/sqrt2=0`