The general solution of the differential equation, y'+y φ'(x)-φ(x) ⋅ φ'(x)=0, where φ(x) is a known function, is where c is an arbitrary constant

Question

The general solution of the differential equation, y'+y φ'(x)-φ(x) ⋅ φ'(x)=0, where φ(x) is a known function, is where c is an arbitrary constant (A) y=c e-φ(x)+φ(x)-1 (B) y=c e+φ(x)+φ(x)-1 (C) y=c e-φ(x)-φ(x)+1 (D) y=c e-φ(x)+φ(x)+1

Tardigrade Admin · Accepted Answer

(d y/d x)+y φ'(x)=φ(x) φ'(x) I.F. =e∫ φ'(x) d x=eφ(x) Hence, the solution is y eφ(x)=∫ eφ(x) φ(x) φ'(x) d x =∫ et t d t, where φ(x)=t =t et-et+c =φ(x) eφ(x)-eφ(x)+c ∴ y=c e-φ(x)+φ(x)-1

Q. The general solution of the differential equation, $y^{'} + y ϕ^{'} (x) - ϕ (x) \cdot ϕ^{'} (x) = 0,$ where $ϕ (x)$ is a known function, is
where $c$ is an arbitrary constant

$y = c e^{- ϕ (x)} + ϕ (x) - 1$

$y = c e^{+ ϕ (x)} + ϕ (x) - 1$

$y = c e^{- ϕ (x)} - ϕ (x) + 1$

$y = c e^{- ϕ (x)} + ϕ (x) + 1$

Q. The general solution of the differential equation, y′+yϕ′(x)−ϕ(x)⋅ϕ′(x)=0, where ϕ(x) is a known function, is where c is an arbitrary constant

y=ce−ϕ(x)+ϕ(x)−1

y=ce+ϕ(x)+ϕ(x)−1

y=ce−ϕ(x)−ϕ(x)+1

y=ce−ϕ(x)+ϕ(x)+1

dxdy​+yϕ′(x)=ϕ(x)ϕ′(x) I.F.=e∫ϕ′(x)dx=eϕ(x) Hence, the solution is yeϕ(x)=∫eϕ(x)ϕ(x)ϕ′(x)dx =∫ettdt, where ϕ(x)=t =tet−et+c =ϕ(x)eϕ(x)−eϕ(x)+c ∴y=ce−ϕ(x)+ϕ(x)−1

Q. The general solution of the differential equation, $y^{'} + y ϕ^{'} (x) - ϕ (x) \cdot ϕ^{'} (x) = 0,$ where $ϕ (x)$ is a known function, is
where $c$ is an arbitrary constant

$y = c e^{- ϕ (x)} + ϕ (x) - 1$

$y = c e^{+ ϕ (x)} + ϕ (x) - 1$

$y = c e^{- ϕ (x)} - ϕ (x) + 1$

$y = c e^{- ϕ (x)} + ϕ (x) + 1$