Find the General solution using variation of parameters yyta

Find the General solution using variation of parameters.

y\'\'+y\'=tanx given that y_c= C₁cosx+C₂sinx

Solution

Given that

y\'\' + y = tanx

The auxialary equation is ,

m² + 1 = 0

  m² = -1

   m = (-1)^1/2

   m = ±i

The roots are imaginary

Hence,

Complementary function is ,y_c = c₁cosx + c₂sinx

y₁(x) = cosx , y₂(x) = sinx

^{Compute the wronskian}

   W(x) = det of y₁ y₂

y₁\' y₂\'

   W(x) = ( y₁y₂\' - y₂y₁\' )

= ( cosx.(cosx) - sinx.(-sinx) )

= cos2x + sin2x

= 1

  W(x) = 1

Here, g(x) = tanx

Compute u₁(x) = - ( y₂(x).g(x) )/W(x) dx

   = - (sinx.tanx) / 1dx

   ^{= - [ ln[(1/cosx ) + tanx ] -sinx ]}

   u₁(x) =  - [ ln[(1/cosx ) + tanx ] + sinx

u₂(x) = ( y₁(x).g(x) )/W(x) dx

   =   ( cosx.tanx ) / 1 dx

= cosxtanx dx

= cosx.(sinx/cosx) dx

   =   sinx dx

   u₂(x) = - cosx

Hence,

   Perticular solution is, y_p(x) = u₁(x).y₁(x) + u₂(x).y₂(x)

   = { - [ ln[(1/cosx ) + tanx ] + sinx }.cosx + (-cosx).sinx

= - cosx.ln(secx + tanx ) +sinxcosx - sinxcosx

= -cosx.ln(secx + tanx )

General solution is ,

y(x) = y_c + y_p

= c₁cosx + c₂sinx + ( - cosx.ln(secx + tanx ) )

   y(x) =c₁cosx + c₂sinx - cosx.ln(secx + tanx )

Therefore,

The general solution is ,

y(x) =c₁cosx + c₂sinx - cosx.ln(secx + tanx )