Question Number 105039 by bemath last updated on 25/Jul/20

(1) ∫ (dx/(x^6 −64))  (2) y′ + y tan x = sin 2x  y(0) = 1

Answered by bobhans last updated on 25/Jul/20

(2) If u(x) = e^(∫ tan (x) dx)  = e^(ln ((1/(cos x))) ) =(1/(cos x))=sec x  y(x) = ((∫ sin 2x.sec x dx + C)/(sec x))  y(x) = cos x.{∫2sin x dx + C }  y(x) = cos x. {−2cos x + C }  y(x) = −2cos ^2 x + C.cos x ★  y(0) = −2+C = 1 ⇔C = 3  y(x) = −2cos ^2 x+3cos x

Answered by mathmax by abdo last updated on 25/Jul/20

1) complex method  we decompose F(z) =(1/(z^6 −64)) ⇒  z^6  =64   let z =re^(iθ)      so e ⇒r^6  e^(i6θ)  =64 e^(i(2kπ))  ⇒r =^6 (√(64)) =2  θ =((kπ)/3)  and k∈[[0,5]] so the roots are z_k =2e^(i((kπ)/3))  and 0≤k≤5  ⇒F(z) =(1/(Π_(k=0) ^5 (z−z_k ))) =Σ_(k=0) ^5  (a_k /(z−z_k ))  with a_k =(1/(6z_k ^5 )) =(z_k /(6(64))) =(z_k /(384))  ⇒F(z) =(1/(384)) Σ_(k=0) ^5  (z_k /(z−z_k )) ⇒∫ F(z)dz =(1/(384)) Σ_(k=0) ^5 z_k  ∫  (dz/(z−2e^((ikπ)/3) ))  =(1/(384)) Σ_(k=0) ^5  (2e^((ikπ)/3) )ln(z−2e^((ikπ)/3) )  +C

Commented byDwaipayan Shikari last updated on 25/Jul/20

I think it can be done with (1/(16))∫(1/(x^3 −8))−∫(1/(x^3 +8))  =(1/(16))∫(A/(x−2))+((Bx+C)/(x^2 +2x+4))−(1/(16))∫((A′)/(x+2))+((B′x+C′)/(x^2 −2x+4))

Commented bymathmax by abdo last updated on 26/Jul/20

yes its another way for tbis integral

Answered by mathmax by abdo last updated on 25/Jul/20

y^′  +ytanx =sin(2x)  h)→y^′  =−ytanx ⇒(y^′ /y) =−tanx ⇒ln∣y∣ =−∫ ((sinx)/(cosx))dx =ln∣cosx∣ +c ⇒  y(x) =k ∣cosx∣  let find solution on {x /cosx >0}  mvc method →y^′  =k^′  cosx −ksinx   e⇒k^′  cosx−ksinx +kcosx ×((sinx)/(cosx)) =sin(2x) ⇒k^′  =((sin(2x))/(cosx)) =2sinx ⇒  k =2∫sinx dx =−2cosx +λ ⇒y(x) =(−2cosx +λ)cosx  ⇒y(x) =λcosx−sin(2x)  y(0)=1 ⇒λ−0 =1 ⇒λ =1 ⇒y(x) =cosx −sin(2x)

Answered by Dwaipayan Shikari last updated on 25/Jul/20

(dy/dx)+ytanx=sin2x  I. F=e^(∫tanxdx) =e^(log(secx)) =secx  y.secx=∫sin2xsecxdx=2∫sinx=−2cosx+C  y=−2cos^2 x+C cosx  y(x)=−2cos^2 x+Ccosx  y(0)=−2+C  C=3  y(x)=−2cos^2 x+3cosx★

Answered by OlafThorendsen last updated on 25/Jul/20

(1) R(x) = (1/(x^6 −1))  R(x) = (1/((x−1)(x+1)(x^2 −x+1)(x^2 −x+1)))  ∫R(x)dx =  ∫((1/(6(x−1)))−(1/(6(x+1)))+((x−2)/(6(x^2 −x+1)))−((x+2)/(6(x^2 +x+1))))dx  = (1/6)ln∣((x−1)/(x+1))∣  +(1/(12))∫((2x−1)/(x^2 −x+1))dx−(1/(12))∫((2x+1)/(x^2 +x+1))dx  −(1/4)∫(dx/(x^2 −x+1))−(1/4)∫(dx/(x^2 +x+1))  =(1/6)ln∣((x−1)/(x+1))∣+(1/(12))ln∣((x^2 −x+1)/(x^2 +x+1))∣  −(1/3)∫(dx/((4/3)(x−(1/2))^2 +1))−(1/3)∫(dx/((4/3)(x+(1/2))^2 +1))  =(1/6)ln∣((x−1)/(x+1))∣+(1/(12))ln∣((x^2 −x+1)/(x^2 +x+1))∣  −(1/(2(√3)))arctan[(2/(√3))(x−(1/2))]−(1/(2(√3)))arctan[(2/(√3))(x+(1/2))]  arctanu+arctanv = arctan((u+v)/(1−uv))  u = (2/(√3))(x−(1/2)) and v = (2/(√3))(x+(1/2))  ((u+v)/(1−uv)) = (((4/(√3))x)/(1−(4/3)(x^2 −(1/4)))) = (((√3)x)/(1−x^2 ))  Finally :  ∫R(x)dx = (1/6)ln∣((x−1)/(x+1))∣+(1/(12))ln∣((x^2 −x+1)/(x^2 +x+1))∣−(1/(2(√3)))arctan(((√3)x)/(1−x^2 ))+C

Commented bybemath last updated on 27/Jul/20

the question ∫ (dx/(x^6 −64)) sir

Answered by OlafThorendsen last updated on 25/Jul/20

Q2.  y′cosx−(−ysinx) = cosxsin2x = 2cos^2 xsinx  ((y′cosx−(−ysinx))/(cos^2 x)) = 2sinx  (d/dx)((y/(cosx))) = 2sinx  (y/(cosx)) = −2cosx+C  y = −2cos^2 x+Ccosx  y(0) = −2+C = 1 ⇒ C = 3  y = 3cosx−2cos^2 x

Answered by bramlex last updated on 26/Jul/20

(1) ∫ (dx/(x^6 −(2)^6 )) = (1/(24))ln ((((x−2)^2 (x^2 −2x+4))/((x+2)^2 (x^2 +2x+4))))−(1/(4(√3)))tan^(−1) ((2/(x^2 +2)))+C