Question

Solve:
(y + z) dx + (z + x) dy + (x + y) dz=0​

Answer 1

Answer:

(x*y + z*x + z*y) = 0  

Step-by-step explanation:

Given differential  equation : ( y + z)dx + (z+x)dy + (x +y)dz = 0

opening the bracket we get :

⇒ y×dx + z×dx +z×dy +x×dy + x×dz + y×dz = 0 ------------------ equation 1

Now integrating both side we get

⇒ ∫y×dx +  ∫z×dx + ∫z×dy + ∫x×dy + ∫x×dz + ∫y×dz = ∫0

we know that : ∫x^n×dx = x^n+1÷n+1

∫x*dy = x∫1*dy = x∫y^0*dy = x*y

using above formula we get

⇒ y*x + z*x + z*y + x*y + x*z+ y*z = 0

⇒ 2(x*y + z*x + z*y) = 0 ⇒ x*y + z*x + z*y = 0

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Answer 2

Solve the differential equation:

(y + z) dx + (z + x) dy + (x + y) dz=0​

opening the bracket of the equation:

ydx + zdx + zdy + xdy + xdz + ydz = 0

Now, integrating the differential equation in both LHS and RHS

∫ydx +  ∫zdx + ∫zdy + ∫xdy + ∫xdz + ∫ydz = ∫0

solving for LHS:

∫ydx =  y∫dx = yx

∫zdx =  z∫dx = zx

∫zdy =  z∫dy = zy

∫xdy =  x∫dy = xy

∫xdz =  x∫dz = xz

∫ydz =  y∫dz = yz

yx + zx + zy + xy + xz + yz  = 0

2( xy + zy +zx) = 0

=> xy + zy +zx = 0

Hence, xy + zy +zx = 0

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