Tentukan solusi dari persamaan diferensial berikut:
dy/dx = (〖4x〗^(2 )+ 〖3y〗^2)/2xy
Jawab:
dy/dx = (〖4x〗^(2 )+ 〖3y〗^2)/2xy
(2xy)dy = (〖4x〗^(2 )+ 〖3y〗^2 ) dx
2x(vx) dy = (〖4x〗^(2 )+ 3 〖(vx)〗^2 ) dx
(〖2x〗^(2 ) v) dy = (4x^2 +3v^2 x^2 ) dx
(2x^2 v) (v dx +x dx)=4x^2 dx +3v^2 x^2 dx
〖2x〗^2 v^2 dx +2x^3 v dv =4x^2 dx +3v^2 x^2 dx
〖2x〗^3 v dv =4x^2 dx +3v^2 x^2 dx-2x^2 v^2 dx
2x^3 v dv =4x^2 dx + v^2 x^2 dx
2x^3 v dv = x^2 (4 + v^2 ) dx
∫▒(2v dv)/(4 + v^(2 ) ) = ∫▒1/x dx
Misalkan:
4 +v^2 =u
2v dv =du
Jadi,
∫▒2v/〖4 +u〗^2 du =∫▒du/u = ln〖u 〗
=ln|4+v^2 |
Jadi integral dari ∫▒(2v dv)/(4 + v^(2 ) ) = ∫▒1/x dx
ln|4+v^2 |=ln|x|+ln〖c 〗
ln〖c =ln 〗 ((4+v^2))/x
c = (4 +v^2)/x
c=(4 +〖y/x〗^2)/x
c=( 4 x^2+ y^2)/x^3
dy/dx= (3y^(3 )- x^3)/〖3xy〗^2
Jawab:
dy/dx= (3y^(3 )- x^3)/〖3xy〗^2
(〖3xy〗^2 )dy=(〖3y〗^3- x^3 )dx
3x (vx)^2 dy= 〖(3(vx)〗^3- x^3) dx
3x 〖(v〗^2 x^2)(vdx+xdv)=(3(v^3 x^3 )- x^3 )dx
(〖3x〗^3 v^2 )(vdx+xdv)=(3 v^3 x^3- x^3 )dx
3x^3 v^3 dx+3x^4 v^2 dv=3x^3 v^3 dx- x^(3 ) dx
3x^4 v^2 dv= -x^3 dx
3v^2 dv= (-x^3 dx)/(x^4 )
3v^2 dv= - 1/x dx
∫▒〖3v^(2 ) dv= ∫▒〖-1/x〗〗 dx
v^3+ c= -ln〖|x|+ln〖c 〗 〗
ln〖|x|+v^3=ln〖c 〗 〗
v^3=ln〖c/x〗
c/x= e^(v^3 )
c=xe^(v^3 )
c=xe^(〖(y⁄(3 ))〗^3 )
dy/dx= y/x-cot(x/y)
Jawab:
dy/dx= y/x-cot(x⁄y)
dy= (y/x- cot〖y/x〗 )dx
v dx+x dv= (v- cotv )dx
v dx+x dv= vdx- cot〖v dx〗
x dv= -cot〖 v 〗 dx
dv/cotv = -dx/x
∫▒dv/cotv = -∫▒dx/x × 1/((x)(cot〖v)〗 )
∫▒(arc v dv)/(〖sin 〗v dv)= -∫▒dx/x
ln〖(sin〖v)= -ln(x)+ln〖(c)〗 〗 〗
ln〖c=ln〖 (sin〖v)-ln(x) 〗 〗 〗
c= (sin x/y)/x
cx= sin x/y
dy/dx= (y- √(x^2-y^2 ))/x
Jawab:
dy/dx= (y- √(x^2-y^2 ))/x
x dy=(y- √(x^2-y^2 ))dx
x dy=y dx- √(x^2-y^2 ) dx
x(v dx+x dv)= vx dx- √(x^2-〖(vx〗^2)) dx
x(v dx+x dv)= vx dx- √(x^2-v^2 x^2 ) dx
xv dx+ x^2 dv = vx dx- √(x^2 (1-v^(2 ) ))dx
dv=-x√((1-v^2 ))dx 〖× 〗〖1/((x^2)√((1-v^2 ))〗
dv/√(1-v^2 )= (-x)/x^2 dx
∫▒dv/√(1-v^2 )= ∫▒〖(-1)/x dx〗
∫▒dv/√(1-v^2 )= -∫▒〖1/x dx〗
-arc cos〖 v=〗-ln〖(x)+ ln(c)〗
arc cos〖 v=〗 ln〖(x)- ln(c)〗
arc cos〖 v=〗 ln x/c
arc cos〖y/x〗=ln x/c
x/c= e^(arc cosv )
c=〖xe〗^(arc cos x/y)
dy/dx=1- y/x- 〖cos〗^2 (y⁄x)
Jawab:
dy/dx=1- y/x- 〖cos〗^2 (y⁄x)
dy/dx=1- v - 〖cos〗^2 (v)
vdx+xdv=(1-v- 〖cos〗^2 (v) )dx
xdv=dx-vdx-vdx- 〖cos〗^2 vdx
xdv=dx-2vdx- 〖cos〗^2 vdx
xdv=(1-2v-〖cos〗^2 v)dx
1/((1-2v-〖cos〗^2 v)) dv= 1/x dx
∫▒〖1/((1-2v-〖cos〗^2 v)) dv〗=∫▒〖1/x dx〗
〖ln 〗〖(1-2v-〖cos〗^2 v〗)+ c=ln〖(x)+ln(c) 〗
〖(1-2v-〖cos〗^2 v〗)=cx
c=(1-2v-〖cos〗^2 v)/x
c=(1-2v+1/2(1+cos2v))/x
c=(1-2v+1/2(1+cos2v))/x
c= (1-2(y/x)+ 1/2+1/2 cos〖2 y/x〗)/x
cx= 1-2(y/x)+ 1/2+1/2 cos〖2 y/x〗
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