Question Number 103767 by bramlex last updated on 17/Jul/20

solve y′−y = y^4 at y(0) = 1

Answered by bemath last updated on 17/Jul/20

substitute v = y^(−3) (dv/dx) = −3y^(−4) (dy/dx) ⇒(dy/dx) = −(y^4 /3) (dv/dx) ⇔ −(y^4 /3) (dv/dx) −y = y^4 −(1/3) (dv/dx)−v = 1 ⇒(dv/dx) +3v = −3 integrating factor u(x)=e^(∫3 dx) u(x)= e^(3x) so we have v = ((∫−3e^(3x) +C)/e^(3x) ) v = ((−e^(3x) +C)/e^(3x) ) = Ce^(−3x) −1 (1/y^3 ) = Ce^(−3x) −1 . y(0)=1 1 = C.1−1 ⇒ C = 2 therefore y^3 = (1/(2e^(−3x) −1))

Answered by mathmax by abdo last updated on 18/Jul/20

y^′ −y =y^4 and y(o)=1 e⇒(y^′ /y^4 )−(1/y^3 ) =1 let (1/y^3 ) =z ⇒z^′ =−((3y^2 y^′ )/y^6 ) =−((3y^′ )/y^4 ) ⇒(y^′ /y^4 ) =−(1/3)z^′ e⇒−(1/3)z^′ −z =1 ⇒(1/3)z^′ +z =−1 ⇒z^′ +3z =−3 h→z^(′ ) =−3z ⇒(z^′ /z) =−3 ⇒ln∣z∣ =−3x +c ⇒z =k e^(−3x) mvc method →z^′ =k^′ e^(−3x) −3k e^(−3x) e ⇒k^′ e^(−3x) −3k e^(−3x) +3k e^(−3x) =−3 ⇒k^′ =−3 e^(3x) ⇒k =−e^(3x) +λ ⇒ z(x) =(−e^(3x) +λ)e^(−3x) =−1 +λe^(−3x) y^3 =(1/z) ⇒y =^3 (√(1/z))=(1/((^3 (√(−1+λe^(−3x) ))))) y(o) =1 ⇒(1/((^3 (√(−1+λ))))) =1 ⇒−1+λ =1 ⇒λ =2 ⇒ y(x) =(1/((^3 (√(−1+2e^(−3x) )))))