solve the differential equation........

(dy/dx)=
(tan(2x+y+3))^2

• let 2x+y+3=z
  => 2+dy/dx = dz/dx
  => dy/dx = (dz/dx -2)
  so eqn becomes
  dz/dx -2 = tan^2z
  => dz/dx= tan^2z+ 2
  => dz/(tan^2z +2)=dx
  let tanz=k then sec^2zdz=dk => dz=dk/(1+k^2)
  => dk/(k^2+1)(k^2+2)=dx
  => [1/(k^2+1)-1/(k^2+2)]dk = dx
  integrate => tan−1(k)- 1/sqrt(2)* tan-(k/sqrt2)=x
  => tan-(tan(2x+y+3))-tan-1(tan(2x+y+3)/sqrt2)=x
  => x+y+3-tan-1(tan(2x+y+3)/sqrt2)=0
  
• 10 years agoHelpfull: Yes(0) No(1)