Question

If $\varphi \left(x\right)$ is a differential function, then the solution of the differential equation $dy+\left\{y{\varphi }^{\prime }\right(x)-\varphi (x\left){\varphi }^{\prime }\right(x\left)\right\}dx=0$ is

• A. $y=\varphi \left(x\right)-1+c{e}^{-\varphi \left(x\right)}$
• B. $y=\varphi \left(x\right)+1+c{e}^{-\varphi \left(x\right)}$
• C. $y{e}^{\varphi \left(x\right)}=\varphi \left(x\right)e^{\varphi \left(x\right)}+c$
• D. $y=\varphi \left(x\right)+1+c{e}^{\theta \left(x\right)}$

Answer 1

The correct option is A $y=\varphi \left(x\right)-1+c{e}^{-\varphi \left(x\right)}$

Given,

$dy+\left(y{\varphi }^{\prime }\right(x)-\varphi (x\left){\varphi }^{\prime }\right(x\left)\right)dx=0$

or

$\frac{dy}{dx}+y{\varphi }^{\prime }\left(x\right)=\varphi \left(x\right){\varphi }^{\prime }\left(x\right)$

$P={\varphi }^{\prime }\left(x\right),q=\varphi \left(x\right).{\varphi }^{\prime }\left(x\right)$

I.F.=$e^{\int p.dx}$

=$e^{{\varphi }^{\prime }\left(x\right)dx}$

=$e^{\varphi \left(x\right)}$

Complete solution

$y\times e^{\varphi \left(x\right)}=\int e^{\varphi \left(x\right)}\times \varphi \left(x\right){\varphi }^{\prime }\left(x\right)dx+c$

$y\times e^{\varphi \left(x\right)}=\int e^{\varphi \left(x\right)}\varphi \left(x\right){\varphi }^{\prime }\left(x\right)dx+c$

Let $\varphi \left(x\right)=t$

${\varphi }^{\prime }\left(x\right)dx=dt$

Put this in above equation

$\therefore y\times e^t=\int e^{t}tdt+c$

$\to y\times e^t=t{e}^t-e^t+c$

$\to y=\varphi \left(x\right)-1+c^{-\varphi \left(x\right)}$