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Question Number 2050 by Yozzi last updated on 01/Nov/15

Solve the d.e (dy/dx)−y−3y^2 =−2 by letting y=((−1)/(3u))((du/dx)).

Solvethed.edydx−y−3y2=−2bylettingy=−13u(dudx).

Answered by 123456 last updated on 01/Nov/15

y′=(dy/dx) u′=(du/dx) y=−((u′)/(3u)) y′=−((3u′′u−3(u′)^2 )/(9u^2 ))=(((u′)^2 −u′′u)/(3u^2 )) (((u′)^2 −u′′u)/(3u^2 ))+((u′)/(3u))−(((u′)^2 )/(3u^2 ))=−2 ((u′)/u)−((u′′)/u)=−6 u′−u′′=−6u u′′−u′−6u=0 −−−−−−−−− u=ce^(λx) λ^2 −λ−6=0 Δ=(−1)^2 −4(1)(−6)=1+24=25 λ=((1±5)/2) u=C_1 e^(−2x) +C_2 e^(3x) y=((2C_1 e^(2x) −3C_2 e^(3x) )/(C_1 e^(−2x) +C_2 e^(3x) ))

y′=dydxu′=dudxy=−u′3uy′=−3u″u−3(u′)29u2=(u′)2−u″u3u2(u′)2−u″u3u2+u′3u−(u′)23u2=−2u′u−u″u=−6u′−u″=−6uu″−u′−6u=0−−−−−−−−−u=ceλxλ2−λ−6=0Δ=(−1)2−4(1)(−6)=1+24=25λ=1±52u=C1e−2x+C2e3xy=2C1e2x−3C2e3xC1e−2x+C2e3x

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