Question

(D2 + 4) y = sec 2x​

Answer 1

Answer:

Notice, C.F. (complementary function) & P.I. (particular integral) of second order linear differential equation: (D2+4)y=tan2x=f(x) is given as

D2+4=0⟹D=±2i

C.F.=C1cos2x+C2sin2x≡C1u(x)+C2v(x)

P.I.=Acos2x+Bsin2x

Where,

A=−∫f(x)v(x) dxu(x)ddxv(x)−v(x)ddxu(x)

=−∫tan2xsin2x dxcos2xddx(sin2x)−sin2xddx(cos2x)

=−∫sin22xsec2x dxcos2x(2cos2x)−sin2x(−2sin2x)

=−12∫(1−cos22x)sec2x dx

=−12∫(sec2x−cos2x)dx

=−12(12ln(sec2x+tan2x)−sin2x2)

=−14(ln(sec2x+tan2x)−sin2x)

B=∫f(x)u(x) dxu(x)ddxv(x)−v(x)ddxu(x)

=∫tan2xcos2x dxcos2xddx(sin2x)−sin2xddx(cos2x)

=∫sin2x dxcos2x(2cos2x)−sin2x(−2sin2x)

=12∫sin2x dx

=12(−cos2x2)

=−cos2x4

setting the values of A & B in above P.I. we get complete solution of given differential equation as follows

y=C.F.+P.I.

y=C1cos2x+C2sin2x−14(ln(sec2x+tan2x)−sin2x)cos2x−cos2x4sin2x

y=C1cos2x+C2sin2x−cos2x4ln(sec2x+tan2x)