We have (D2+a2)y=secaxA.E. is m2+a2=0⇒m=±aiC.F. =yc=C1cosax+C2sinaxy=Acosax+Bsinax…(1) be the complete solution of the given equation where A and B are to be found.We have, y1=cosax,y2=sinaxy1′=−asinax,y2′=acosaxW=y1y2′−y2y1′=a. Also, ϕ(x)=secaxA′=W−y2ϕ(x), and B′=Wy1ϕ(x)A′=a−sinax⋅secax,B′=acosax⋅secaxA′BAA=a−tanax,B′=a1=a1∫dx+c2=−a1∫tanaxdx+C1,B=ax+C2=a2−log(secax)+C1Substituting these values of A and B in Eqn. (1), we get:
y=C1cosax+C2sinax−a2cosaxlog(secax)+axsinax
Comments