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Question Number 80293 by ~blr237~ last updated on 01/Feb/20
Find all functions that  satisfy to    (E): ∀ x∈R      xf(x)+∫_0 ^x f(x−t)cos(2t)dt=sin(2x)
Findallfunctionsthatsatisfyto(E):xRxf(x)+0xf(xt)cos(2t)dt=sin(2x)
Commented by john santu last updated on 02/Feb/20
2y+xy′ = 2cos 2x ⇒ differential  equation
2y+xy=2cos2xdifferentialequation
Commented by ~blr237~ last updated on 02/Feb/20
just check your mistake on the (3−4)lines  crossing
justcheckyourmistakeonthe(34)linescrossing
Commented by john santu last updated on 02/Feb/20
(d/dx)[∫_0 ^x f(u)cos (x−u)du]=  f(x)cos (x)+f(0) cos (0) ?
ddx[x0f(u)cos(xu)du]=f(x)cos(x)+f(0)cos(0)?
Commented by ~blr237~ last updated on 02/Feb/20
No
No
Commented by jagoll last updated on 02/Feb/20
what the right?
whattheright?
Commented by ~blr237~ last updated on 02/Feb/20
like that....?what is not valid? be clear?
likethat.?whatisnotvalid?beclear?
Commented by jagoll last updated on 02/Feb/20
we know that  ∫_a ^x  f(u)du ⇒ (d/dx) ∫_a ^x f(u)du= f(x)  it clear by fundamental  calculus 1.   so (d/dx)∫_1 ^( x) f(u)cos (x−u)du =   f(x) cos (x−x) = f(x)
weknowthataxf(u)duddxaxf(u)du=f(x)itclearbyfundamentalcalculus1.soddx1xf(u)cos(xu)du=f(x)cos(xx)=f(x)
Commented by ~blr237~ last updated on 02/Feb/20
sir what is going  on?   let take f(t)=t  F(x)=∫_0 ^x (x−t)cos2tdt         by part   u′=cos2t  and v=x−t  F(x)=[(1/2)(x−t)sin2t]_0 ^x  +(1/2)∫_0 ^x sin2tdt    =(1/2)[((−cos2t)/2)]_0 ^x =((1−cos2x)/4)  (dF/dx)=((sin2x)/2) ≠ f(x)=x
sirwhatisgoingon?lettakef(t)=tF(x)=0x(xt)cos2tdtbypartu=cos2tandv=xtF(x)=[12(xt)sin2t]0x+120xsin2tdt=12[cos2t2]0x=1cos2x4dFdx=sin2x2f(x)=x
Commented by jagoll last updated on 02/Feb/20
okay sir
okaysir
Commented by ~blr237~ last updated on 02/Feb/20
When you want to derivate F(x)=∫_0 ^(u(x))  f(x,t)dt make  sure firstly (by a good stating variable: may be  z=(t/(u(x)))  ) to remove x at the whole boundary  we were having  F(x)=∫_0 ^x f(x−t)cos2tdt  F(x)=∫_0 ^1 x[f(x−xv)cos(2xv)]dv   with  v=(t/x)   Now you can use  (dF/dx)=∫_0 ^1 (∂/∂x) g(x,v)dv       where  g(x,v)=xf(x−xv)cos(2xv)
WhenyouwanttoderivateF(x)=0u(x)f(x,t)dtmakesurefirstly(byagoodstatingvariable:maybez=tu(x))toremovexatthewholeboundarywewerehavingF(x)=0xf(xt)cos2tdtF(x)=01x[f(xxv)cos(2xv)]dvwithv=txNowyoucanusedFdx=01xg(x,v)dvwhereg(x,v)=xf(xxv)cos(2xv)
Commented by mr W last updated on 02/Feb/20
to jagoll sir:  ∫_a ^x  f(u)du ⇒ (d/dx) ∫_a ^x f(u)du= f(x)  is correct. but it can not be applied to  ∫_a ^x  f(u,x)du ⇒ (d/dx) ∫_a ^x f(u,x)du≠ f(x,x)
tojagollsir:axf(u)duddxaxf(u)du=f(x)iscorrect.butitcannotbeappliedtoaxf(u,x)duddxaxf(u,x)duf(x,x)
Commented by mr W last updated on 02/Feb/20

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