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Question Number 106774 by bemath last updated on 07/Aug/20

     ^(@bemath@)    (1)    3^x  + 3^((√x) ) = 90. find x ?     (2) x (dy/dx)−(1+x)y = xy^2

@bemath@(1)3x+3x=90.findx?(2)xdydx(1+x)y=xy2

Answered by john santu last updated on 07/Aug/20

(1) by observe x = 4 ⇒3^(4 )  + 3^(√4)  =  81 + 9 = 90. ^(@JS@)

(1)byobservex=434+34=81+9=90.@JS@

Answered by bobhans last updated on 07/Aug/20

        ^(≻bobhans≺)   (dy/dx) − (((1+x)/x))y = y^2   (Bernoulli diff eq )  set u = y^(−1)  ⇒(du/dx) = −y^(−2)  (dy/dx) ; (dy/dx) =−y^2  (du/dx)  ⇔ −y^2  (du/dx)−(((1+x)/x))y=y^2   (du/dx) + (((1+x)/x)) u = −1 ; integrating factor   v(x)= e^(∫ (1+(1/x))dx) = e^(x+ln x)  = xe^x   u(x)= ((∫−xe^x  dx +C)/(xe^x )) = ((−xe^x +e^x +C)/(xe^x ))  (1/y) = ((−xe^x +e^x +C)/(xe^x )) ⇒ y = ((xe^x )/(C−xe^x +e^x ))

bobhansdydx(1+xx)y=y2(Bernoullidiffeq)setu=y1dudx=y2dydx;dydx=y2dudxy2dudx(1+xx)y=y2dudx+(1+xx)u=1;integratingfactorv(x)=e(1+1x)dx=ex+lnx=xexu(x)=xexdx+Cxex=xex+ex+Cxex1y=xex+ex+Cxexy=xexCxex+ex

Answered by abdomathmax last updated on 07/Aug/20

3) xy^′ −(1+x)y =xy^2  ⇒(y^′ /y^2 )−(1+(1/x))(1/y)=1  let z =(1/y) ⇒z^′  =−(y^′ /y^2 )  e⇒−z^′ −(1+(1/x))z =1 ⇒z^′  +(1+(1/x))z =−1  he→z^′ =−(1+(1/x))z ⇒(z^′ /z) =−(1+(1/x)) ⇒  ln∣z∣ =−x−ln∣x∣ +c ⇒z = k e^(−x) ×(1/(∣x∣))  solution on{x /x>0} ⇒z =k (e^(−x) /x)  mvc mrthod→z^(′ )  =k^′  (e^(−x) /x) +k ×((−xe^(−x) −e^(−x) )/x^2 )  =k^′  (e^(−x) /x) −k  (e^(−x) /x) −k (e^(−x) /x^2 )  e ⇒ k^′  (e^(−x) /x) −k (e^(−x) /x)−k (e^(−x) /x^2 ) +(1+(1/x))k (e^(−x) /x) =−1  ⇒k^′   =−x e^x  ⇒k =−∫x e^x  dx +c  =−{ xe^x −∫ e^x  dx} +c=−(x−1)e^x  +c ⇒  z =(e^(−x) /x)( −(x−1)e^x +c) =−((x−1)/x) +((ce^(−x) )/x)  we hsvr y =(1/z) ⇒ y =(x/(1−x +ce^(−x) ))

3)xy(1+x)y=xy2yy2(1+1x)1y=1letz=1yz=yy2ez(1+1x)z=1z+(1+1x)z=1hez=(1+1x)zzz=(1+1x)lnz=xlnx+cz=kex×1xsolutionon{x/x>0}z=kexxmvcmrthodz=kexx+k×xexexx2=kexxkexxkexx2ekexxkexxkexx2+(1+1x)kexx=1k=xexk=xexdx+c={xexexdx}+c=(x1)ex+cz=exx((x1)ex+c)=x1x+cexxwehsvry=1zy=x1x+cex

Answered by Dwaipayan Shikari last updated on 07/Aug/20

(dy/dx)−(((1+x))/x)y=y^2                 (1/y) (dy/dx)−((1+x)/x)=y                              {(1/y)=v ,   (dy/dx)v+y(dv/dx)=0  ,(1/y).(dy/dx)=−(1/v).(dv/dx)  −(1/v).(dv/dx)−((1+x)/x)=(1/v)  (dv/dx)+((1+x)/x)v=−1  I.F=e^(∫((1+x)/x)) =e^((logx+x)) =xe^x   v.xe^x =−∫xe^x   v.xe^x =−xe^x +e^x +C  v=−1+(1/x)+(C/x)e^(−x)   (1/y)=−1+(1/x)+(C/x)e^(−x)   (1/y)=((−xe^x +e^x +C)/(xe^x ))  y=((xe^x )/((e^x −xe^x +C)))

dydx(1+x)xy=y21ydydx1+xx=y{1y=v,dydxv+ydvdx=0,1y.dydx=1v.dvdx1v.dvdx1+xx=1vdvdx+1+xxv=1I.F=e1+xx=e(logx+x)=xexv.xex=xexv.xex=xex+ex+Cv=1+1x+Cxex1y=1+1x+Cxex1y=xex+ex+Cxexy=xex(exxex+C)

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