Question

solve ( D2 +4)y = 2 tanx : x > 0 , D = d/dx​

Answer 1

Answer:

We have,

dx

dy

+ytan x=sin x

Comparing with, standard first order linear differential equation

dx

dy

+Py=Q

We get P=tanx and Q=sinx

Thus, integrating factor. I.F=e

∫Pdx

=e

∫tanxdx

=e

lnsecx

=secx

Therefore solution is given by,

y(I.F.)=∫Q(I.F)dx+C

⇒y(secx)=∫secxsinxdx

⇒(secx)y=∫tanxdx+C

⇒(secx)y=ln∣secx∣+C

⇒y=cosxln∣secx∣+Ccosx