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) )))))