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Question Number 155101 by SANOGO last updated on 25/Sep/21

soit: y y′+xy^2 +x=0 ,avec f(0)=1  f′′(0)=?

soit:yy+xy2+x=0,avecf(0)=1f(0)=?

Commented by tabata last updated on 25/Sep/21

yy^′  = −x (y^2  + 1) ⇒ ((y )/(y^2 +1)) dy = −x dx    (1/2) ln ∣ y^2 +1∣ = −(x^2 /2) + (c/2) ⇒ ln ∣ y^2 +1∣ = −x^2 +c    y^2  = k e^(−x^2 ) − 1 ⇒ y = (√(k e^(−x^2 ) −1))    f (0) = 1 ⇒ 1 = (√(k −1)) ⇒ k = 2    ∴ y = (√(2 e^(−x^2 ) −1))    y^′  = ((−  2 x e^(−x^2 ) )/( (√(2e^(−x^2 ) −1)))) ⇒ y^(′′)  = ((((√(2e^(−x^2 ) −1)))(4x^2 e^(−x^2 ) −2 e^(−x^2 ) )− (2 x e^(−x^2 ) )((( 2 x e^(−x^2 ) )/( (√(2 e^(−x^2 ) −1))))))/((2 e^(−x^2 )  − 1)))    f^( ′′) (0) = ((( (√(2 − 1)) ) (− 2 ))/(( 2 − 1 ))) = − (2/( 1)) = −2    < M . T  >

yy=x(y2+1)yy2+1dy=xdx12lny2+1=x22+c2lny2+1=x2+cy2=kex21y=kex21f(0)=11=k1k=2y=2ex21y=2xex22ex21y=(2ex21)(4x2ex22ex2)(2xex2)(2xex22ex21)(2ex21)f(0)=(21)(2)(21)=21=2<M.T>

Commented by SANOGO last updated on 25/Sep/21

merci bien

mercibien

Commented by tabata last updated on 25/Sep/21

you are welcome

youarewelcome

Answered by mr W last updated on 25/Sep/21

y(0)=1  yy′+xy^2 +x=0  1×y′(0)+0×1^2 +0=0  ⇒y′(0)=0  yy′′+(y′)^2 +y^2 +2xyy′+1=0  1×y′′(0)+(0)^2 +1^2 +2×0×1×0+1=0  ⇒y′′(0)=−2

y(0)=1yy+xy2+x=01×y(0)+0×12+0=0y(0)=0yy+(y)2+y2+2xyy+1=01×y(0)+(0)2+12+2×0×1×0+1=0y(0)=2

Commented by SANOGO last updated on 25/Sep/21

merci bien

mercibien

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