If dy = (secx + ytanx)dx, Then the curve

• the curve is sinusoidal in nature.
  y = sinx + k.cosx
  
  expalnation-
  equation is in the form of
  dy/dx -P(x)*y = Q(x)
  so,
  P(x)=-tanx, Q(x)=secx
  Integratiing factor (I.F.) = secx
  after solving this you will get.
  y = sinx + k*cosx, where k = any arbitary cons
• 11 years agoHelpfull: Yes(4) No(1)
• To find the required curve, we have to find y by integrating the given equation as follows:
  Let us convert the given inexact differential equation into an exact differential equation.
  
  Given that,
  dy = [(secx - y) / tanx)]dx
  
  First we have to express the given equation into the form as y' + p(x)y = q(x) ......(1)
  
  i.e, dy = [(secx - y) / tanx)]dx
  dy / dx = (secx - y) / tanx
  dy / dx = secx / tanx - y / tanx
  dy / dx = 1 / sinx - y / tanx
  dy / dx + y/ tanx = cosecx
  dy /dx + ycotx = cosecx .....(2)
  
  Compare (1) and (2), we have p(x) = cotx, q(x) = cosecx
  
  Let us find the integrating factor which can multiply through the eqn(2) in order to bring the left-hand side under a common derivative. For the canonical first-order,
  For the above linear differential equation(1), the integrating factor is e∫p(x)dx
  
  Then, e∫p(x)dx = e∫cot(x) dx
  
  Since ∫cotx dx = log(sinx)
  
  Then, Integrating factor = elog(sinx) = sin(x)
  
  Now, Multiply both sides of eqn2 by the integrating factor then
  
  (sinx) dy / dx + ycotx sinx = cosecx sinx
  (sin x) dy / dx + (cos x) y = 1 which is an exact differential eqn.
  
  d/dx [y(sin x)] = 1 [ By {d / dx[y(sin x)] = (sin x) dy / dx + (cos x) y} by using d(uv)=vdu + udv ]
  Integrating above with respect to x, we get
  
  ysin x = x
  
  y = x / sin x = xcosecx
  
  Hence the curve is y = x(cosecx)
• 10 years agoHelpfull: Yes(0) No(0)
• ans:x=ycosx
  
  as equation is in the form of
  dy/dx -P(x)*y = Q(x)
  
  Q(x)=sec x
  P(x)=tanx
  Integration Factor=cos x
  solution is:y(if)=integration(sec x * IF)+c
  thus
  Y COS x=X +c
  
• 10 years agoHelpfull: Yes(0) No(0)