Question

If $Ï•\left(x\right)$ is a differentiable function, then the solution of the differential equation
$dy+\left\{y{ϕ}^{′}\left(x\right)ϕ\left(x\right){ϕ}^{′}\left(x\right)\right\}dx=0$, is

• A. $y=\left\{Ï•\left(x\right)−1\right\}+C{e}^{−ϕ\left(x\right)}$
• B. $yÏ•\left(x\right)={\left\{Ï•\left(x\right)\right\}}^2+C$
• C. $y{e}^{Ï•\left(x\right)}=Ï•\left(x\right)e^{Ï•\left(x\right)}+C$
• D. $y−ϕ\left(x\right)=Ï•\left(x\right)e^{−ϕ\left(x\right)}$

Answer 1

Answer: (A)

We have,
$dy+\left\{y{ϕ}^{′}\left(x\right)−ϕ\left(x\right){ϕ}^{′}\left(x\right)\right\}dx=0$
$⇒\frac{dy}{dx}+{Ï•}^{′}\left(x\right)â‹\dots y=Ï•\left(x\right){Ï•}^{′}\left(x\right)$
This is a linear differential equation. Here,
$I.F.=e^{∫{ϕ}^{′}\left(x\right)dx}=e^{ϕ\left(x\right)}$
$∴$ Solution is, $y{e}^{ϕ\left(x\right)}=∫$ $\underset{I}{\underbrace{ϕ(x)}}$ $\underset{II}{\underbrace{e^{ϕ\left(x\right)}{ϕ}^{′}\left(x\right)}}dx$
$⇒y{e}^{ϕ\left(x\right)}=ϕ\left(x\right)e^{ϕ\left(x\right)}−∫{ϕ}^{′}\left(x\right)e^{ϕ\left(x\right)}dx$
$⇒y=\left(ϕ\left(x\right)−1\right)+C{e}^{−ϕ\left(x\right)}$.