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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))

(cosx)dydx+ysinx=2xcos2x,y(π4)=15π2232

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 ⊛)

dydx+y.tan(x)=2xcosxIF:u(x)=etanxdx=edcosxcosxu(x)=eln(cosx)=1cosxdycosx.dx+ysinxcos2x=2xddx(ycosx)=2xycosx=x2+Cy(x)=cosx(x2+C)15π2232=22(π216+C)15π216=π216+CC=π2solutiony(x)=(x2π2).cosx(JS)

Commented by bramlex 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

cosxy+sinxy=2xcos2xwithy(π4)=15π232hecosxy+sinxy=0cosxy=sinxyyy=sinxcosxlny=ln(cosx)+cy=kcosxlagrangemethody=kcosxksinxekcos2xkcosxsinx+ksinxcosx=2xcos2xk=2xk(x)=x2+λy(x)=(x2+λ)cosxy(π4)=15π232(π28+λ)×22=15π232λ+π28=15π16λ=π2815π16=17π16y(x)=(x217π16)cosx

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

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

sirx=π4thenx2=π216.?

Commented by mathmax 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

sorryinthequstiony(π4)=15π2232(π216+λ)×22=15π2232π216+λ=15π216λ=π2y(x)=(x2π2)cosx

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