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Question Number 169051 by Mastermind last updated on 23/Apr/22

Solve the ODE  (x^2 −2)y′ + xy = 0, with y(1)=1    Mastermind

SolvetheODE(x22)y+xy=0,withy(1)=1Mastermind

Commented by haladu last updated on 23/Apr/22

      y′ ( x^2 −2 )  =  −xy        (dy/y)  =  −((x dx)/(x^2  −2))    ∫ (dy/y) =  − (1/2) ∫ ((2x)/(x^2  −2))        ln (y)  =  −(1/2) ln (x^2  −2 ) + C     y = x =1     ln (1)  =  −(1/2) ln  (1^2 −2 )  + C    0  =  −(1/2)ln (−1)  + C     C  =  (1/2)ln (−1)     ln (y) = − (1/2)ln (x^2  −2 )  + (1/2)ln (−1)    −2ln (y) =  ln ( x^2 −2 )−ln (−1)         (1/y^2 ) =  ((x^2 −2)/(−1))      y^2   =  2  −x^2

y(x22)=xydyy=xdxx22dyy=122xx22ln(y)=12ln(x22)+Cy=x=1ln(1)=12ln(122)+C0=12ln(1)+CC=12ln(1)ln(y)=12ln(x22)+12ln(1)2ln(y)=ln(x22)ln(1)1y2=x221y2=2x2

Commented by Mastermind last updated on 23/Apr/22

Could you please recheck your solution?

Couldyoupleaserecheckyoursolution?

Commented by haladu last updated on 23/Apr/22

 yes i was in hurry becuse it was time for sallat.

yesiwasinhurrybecuseitwastimeforsallat.

Commented by Mastermind last updated on 24/Apr/22

Okay i understand

Okayiunderstand

Answered by Mathspace last updated on 23/Apr/22

(x^2 −2)y^′ +xy=0 ⇒(x^2 −2)y^′ =−xy ⇒  (y^′ /y)=−(x/(x^2 −2)) ⇒ln∣y∣=−(1/2)ln∣x^2 −2∣ +c  ⇒y=(k/( (√(∣x^2 −2∣))))  solution on w={x/x^2 −2<0}  ⇒y=(k/( (√(2−x^2 ))))=k(2−x^2 )^(−(1/2))   ⇒y^′ =k^′ (2−x^2 )^(−(1/2)) +k((1/2)2x)(x^2 −2)^(−(3/2))   =k^′ (2−x^2 )^(−(1/2)) +xk(2−x^2 )^(−(3/2))   (e)⇒(x^2 −2)k^′ (2−x^2 )^(−(1/2))   −xk(2−x^2 )^(−(1/2))  +xk(2−x^2 )^(−(1/2))   =0⇒k^′ =0 ⇒k=λ ⇒  y(x)=(λ/( (√(2−x^2 ))))  y(1)=1 ⇒λ=1 ⇒y(x)=(1/( (√(2−x^2 ))))

(x22)y+xy=0(x22)y=xyyy=xx22lny∣=12lnx22+cy=kx22solutiononw={x/x22<0}y=k2x2=k(2x2)12y=k(2x2)12+k(122x)(x22)32=k(2x2)12+xk(2x2)32(e)(x22)k(2x2)12xk(2x2)12+xk(2x2)12=0k=0k=λy(x)=λ2x2y(1)=1λ=1y(x)=12x2

Commented by Mastermind last updated on 24/Apr/22

Sir, i think answer is y=(i/( (√(x^2 −2))))

Sir,ithinkanswerisy=ix22

Commented by Mathspace last updated on 24/Apr/22

no

no

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