Question

2)solve (2x^2+2xy+2xz^2+1)dx+dy+2zdz=0

Answer 1

$P=2x^ 2 +2xy+2xz^ 2+1 ,Q=1,R=2z$

If x is a constant $, dx=0,$

$dy+2zdz=0,$

$\int dy+\int 2zdz=C,$

$y+z^2=f(x).$

Differentiate $y+z^2=f(x),$

$dy+2zdz-f'(x)dx=0,$ $f ' (x)=−(2x^ 2 +2xy+2xz^2+1),$ $f'(x) =-(2x^2+1) - 2x( y+z^2 ).$

Replace $y+z^2=f(x),$

$f'(x) =-(2x^2+1)-2xf(x)$,

$f ' (x)+2xf(x)=(−2x ^ 2 −1)$.

Solve the linear equation $f'(x)+2xf(x)=−(2x^ 2 +1),$

$e^{ ∫2xdx}=e^{ x^ 2 },$

$e^{x^2}f(x)=-\int e^{x^2}(2x^2+1)dx+C=$

$-\int x\cdot2xe^{x^2}dx-\int e^{x^2}dx+C=$

$-xe^{x^2}+\int e^{x ^2}dx- \int e^{x^2} dx+C=$

$=−xe ^ { x ^ 2 }+C,$

$f(x) =-x+Ce^{-x^2 },$

$y+z^2=-x+Ce^{-x^2},$

$x+y+z^2=Ce ^ { −x ^ 2 }$