Question Number 101848 by bramlex last updated on 05/Jul/20

(cos x) (dy/dx) +y sin x = 2x cos ^2 x ,  y((π/4)) = ((−15π^2 (√2))/(32))

Answered by john santu last updated on 05/Jul/20

(dy/dx) + y.tan (x)= 2x cos x  ⇒IF : u(x)=e^(∫ tan x dx)  = e^(−∫((dcos x)/(cos x)))   u(x)=e^(−ln (cos x)) =(1/(cos x))  ⇒(dy/(cos x.dx)) + ((y sin x)/(cos ^2 x)) = 2x  (d/dx) ((y/(cos x))) = 2x ⇒(y/(cos x)) = x^2 +C  y(x) = cos x (x^2 +C)  ((−15π^2 (√2))/(32)) = ((√2)/2)((π^2 /(16)) +C)  ⇒−((15π^2 )/(16)) = (π^2 /(16)) +C  C = −π^2   ∴solution y(x)= (x^2 −π^2 ).cos x   (JS ⊛)

Commented bybramlex last updated on 05/Jul/20

macho...😁😁😁

Answered by mathmax by abdo last updated on 05/Jul/20

cosx y^′  +sinx y =2xcos^2 x  with y((π/4)) =−((15π(√2))/(32))  he→cosx y^′  +sinx y =0 ⇒cosx y^′  =−sinx y ⇒(y^′ /y) =−((sinx)/(cosx)) ⇒  ln∣y∣ =ln(cosx)+c ⇒y =k cosx  lagrange method →y^′  =k^′  cosx −ksinx  e⇒k^′  cos^2 x−kcosx sinx +ksinx cosx =2xcos^2 x ⇒k^′  =2x ⇒k(x) =x^2  +λ ⇒  y(x) =(x^2  +λ)cosx   y((π/4)) =−((15π(√2))/(32)) ⇒((π^2 /8) +λ)×((√2)/2) =−((15π(√2))/(32)) ⇒λ +(π^2 /8) =−((15π)/(16)) ⇒  λ =−(π^2 /8)−((15π)/(16)) =−((17π)/(16)) ⇒y(x) =(x^2 −((17π)/(16)))cosx

Commented byjohn santu last updated on 05/Jul/20

sir x = (π/4) then x^2 =(π^2 /(16)) .?

Commented bymathmax by abdo last updated on 05/Jul/20

sorry in the qustion y((π/4))=−((15π^2 (√2))/(32)) ⇒((π^2 /(16))+λ) ×((√2)/2) =−((15π^2 (√2))/(32)) ⇒  (π^2 /(16)) +λ =−((15π^2 )/(16)) ⇒λ =−π^2  ⇒y(x) =(x^2 −π^2 )cosx