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Question Number 12245 by tawa last updated on 16/Apr/17

Solve the differential equation  (dy/dx) = ((4x + 2y − 3)/(8x − 4y + 5))

Solvethedifferentialequationdydx=4x+2y38x4y+5

Answered by mrW1 last updated on 17/Apr/17

(dy/dx) = ((4x + 2y − 3)/(8x − 4y + 5))    let y=u−p  let x=v−q  with p,q=constants  dy=du  dx=dv  4(v−q)+2(u−p)−3=4v+2u+(−4q−2p−3)  8(v−q)−4(u−p)+5=8v−4u+(−8q+4p+5)    −4q−2p−3=0   (i)  −8q+4p+5=0    (ii)  (ii)−(i)×2:  8p+11=0  ⇒p=−((11)/8)  (ii)+(i)×2:  −16q−1=0  ⇒q=−(1/(16))    (dy/dx) = ((4x + 2y − 3)/(8x − 4y + 5)) ⇒  (du/dv) = ((4v + 2u)/(8v − 4u))  let u=wv  (du/dv)=w+v(dw/dv)    (du/dv) = ((4v + 2u)/(8v − 4u))⇒  w+v(dw/dv)=((4+2w)/(8−4w))=((2+w)/(4−2w))  v(dw/dv)=((2+w)/(4−2w))−w=((2−3w+2w^2 )/(4−2w))  ((2−w)/(2w^2 −3w+2))dw=(1/(2v))dv  ∫((2−w)/(2w^2 −3w+2))dw=∫(1/(2v))dv  2∫(1/(2w^2 −3w+2))dw−∫(w/(2w^2 −3w+2))dw=(1/2)ln v+C  (√(4×2×2−9))=(√7)  ∫(1/(2w^2 −3w+2))dw=(2/(√7))tan^(−1) (((4w−3)/(√7)))  ∫(w/(2w^2 −3w+2))dw=(1/4)ln (2w^2 −3w+2)+(3/(2(√7)))tan^(−1) (((4w−3)/(√7)))    (5/(2(√7)))tan^(−1) (((4w−3)/(√7)))−(1/4)ln (2w^2 −3w+2)=(1/2)ln v+C  (5/(√7))tan^(−1) (((4w−3)/(√7)))−(1/2)ln (2w^2 −3w+2)=ln v+C  (5/(√7))tan^(−1) (((4u−3v)/((√7)v)))−(1/2)ln [2((u/v))^2 −3((u/v))+2]=ln v+C  (5/(√7))tan^(−1) [((4(y−((11)/8))−3(x−(1/(16))))/((√7)(x−(1/(16)))))]−(1/2)ln [2(((y−((11)/8))/(x−(1/(16)))))^2 −3(((y−((11)/8))/(x−(1/(16)))))+2]=ln (x−(1/(16)))+C  (5/(√7))tan^(−1) [((4(16y−22)−3(16x−1))/((√7)(16x−1)))]−(1/2)ln [2(((16y−22)/(16x−1)))^2 −3(((16y−22)/(16x−1)))+2]=ln (((16x−1)/(16)))+C  (5/(√7))tan^(−1) [((64y−48x−85)/((√7)(16x−1)))]−(1/2)ln [8(((8y−11)/(16x−1)))^2 −6(((8y−11)/(16x−1)))+2]=ln (((16x−1)/(16)))+C

dydx=4x+2y38x4y+5lety=upletx=vqwithp,q=constantsdy=dudx=dv4(vq)+2(up)3=4v+2u+(4q2p3)8(vq)4(up)+5=8v4u+(8q+4p+5)4q2p3=0(i)8q+4p+5=0(ii)(ii)(i)×2:8p+11=0p=118(ii)+(i)×2:16q1=0q=116dydx=4x+2y38x4y+5dudv=4v+2u8v4uletu=wvdudv=w+vdwdvdudv=4v+2u8v4uw+vdwdv=4+2w84w=2+w42wvdwdv=2+w42ww=23w+2w242w2w2w23w+2dw=12vdv2w2w23w+2dw=12vdv212w23w+2dww2w23w+2dw=12lnv+C4×2×29=712w23w+2dw=27tan1(4w37)w2w23w+2dw=14ln(2w23w+2)+327tan1(4w37)527tan1(4w37)14ln(2w23w+2)=12lnv+C57tan1(4w37)12ln(2w23w+2)=lnv+C57tan1(4u3v7v)12ln[2(uv)23(uv)+2]=lnv+C57tan1[4(y118)3(x116)7(x116)]12ln[2(y118x116)23(y118x116)+2]=ln(x116)+C57tan1[4(16y22)3(16x1)7(16x1)]12ln[2(16y2216x1)23(16y2216x1)+2]=ln(16x116)+C57tan1[64y48x857(16x1)]12ln[8(8y1116x1)26(8y1116x1)+2]=ln(16x116)+C

Commented by tawa last updated on 17/Apr/17

Wow, God bless you sir. I really appreciate sir.

Wow,Godblessyousir.Ireallyappreciatesir.

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