Answer:
y = In | cos(2x + c ) | + c
Step-by-step explanation:
y" + (y')^2 + 4 = 0
substituting u = y'
u' + u^2 + 4 = 0
hence : u' = - (u^2 + 4 )
[tex]\frac{u'}{-(u^2 + 4)}[/tex] = 1 ------- (1)
integrating both sides of the equation 1
[tex]1/2 \int\limits^1_1 {\frac{2du}{(u^2+4)} } \, = x + c[/tex]
x + c = [tex]- \frac{1}{2} arc tan (\frac{u}{2} )[/tex] hence u = -2 tan(2x + c )
remember u = y'
y' = -2 tan(2x + c) ------ (2)
integrating both sides of the equation 2
y = ∫ [tex]\frac{-sin u}{cos u } du[/tex]
therefore Y = In | cosu | + c
y = In | cos(2x + c ) | + c
