Question Number 103553 by byaw last updated on 15/Jul/20

Answered by mathmax by abdo last updated on 16/Jul/20

2y^(′′) −4y^′ −6y =0 ⇒y^(′′) −2y^′  −3y =0  →r^2 −2r +3 =0 →Δ^′  =1+3 =4 ⇒r_1 =1+2 =3 and r_2 =1−2 =−1 ⇒  y =α e^(−x)  +βe^(3x)   y(o)=3 ⇒α+β =3  y^′  =−α e^(−x)  +3β e^(3x)  so y^′ (0) =4 ⇒−α+3β =4 we get the system   { ((α+β =3)),((−α+3β =4 ⇒4β =7 ⇒β =(7/4))) :}  α=3−β =3−(7/4) =(5/4) ⇒y(x) =(5/4)e^(−x) +(7/4)e^(3x)

Answered by bobhans last updated on 16/Jul/20

(3) (∂M/∂y) = −(x/y^2 ).e^(x/y)     (∂N/∂x) = (1/y)e^(x/y) (1−(x/y))−(1/y).e^(x/y)  = −(x/y^2 ).e^(x/y)   because (∂M/∂y) = (∂N/∂x) so this is exact diff  equation

Commented bybyaw last updated on 16/Jul/20

Thank you soo much.

Answered by Dwaipayan Shikari last updated on 16/Jul/20

(dy/dx)=((x^2 +y^2 )/(2xy))  (dy/dx)=((v^2 +1)/(2v))     [vy=x   ,(dy/dx)v+(dv/dx)y  (dv/dx).y=((v^2 +1)/2)  (dv/dx).(x/v)=((v^2 +1)/2)  ((2dv)/(v(v^2 +1)))=(dx/x)  −∫((−(2/v^3 ))/(1+(1/v^2 )))dv=logx+C  −log(1+(1/v^2 ))=logx+C  2logv−log(v^2 +1)=logx+C  2logx−2logy−log(x^2 +y^2 )+2logy=logx+logC_1   log((x/(x^2 +y^2 )).(1/C_1 ))=0  x=C_1 (x^2 +y^2 )

Commented bybyaw last updated on 16/Jul/20

Thank you. I am greatful