Question

Let $\frac{dy}{dx}=\frac{y{\varphi }^{{}^{\prime }}\left(x\right)-y^2}{\varphi \left(x\right)}$, where $\varphi \left(x\right)$ is a specified function satisfying $\varphi \left(1\right)=1,\varphi \left(4\right)=1296$. If $y\left(1\right)=1$ then $\frac{1}{81}y\left(4\right)$ is equal to

Answer 1

Given: $\frac{dy}{dx}=\frac{y{\varphi }^{{}^{\prime }}\left(x\right)-y^2}{\varphi \left(x\right)}$
$\varphi \left(1\right)=1$
$\varphi \left(4\right)=1296$
$y\left(1\right)=1$
$\frac{dy}{dx}=y\frac{{\varphi }^{{}^{\prime }}\left(x\right)}{\varphi \left(x\right)}-\frac{y^2}{\varphi \left(x\right)}$
$\Rightarrow \frac{1}{y^2}\frac{dy}{dx}-\frac{1}{y}\frac{{\varphi }^{{}^{\prime }}\left(x\right)}{\varphi \left(x\right)}=\frac{-1}{\varphi \left(x\right)}$
$\Rightarrow \frac{-\varphi \left(x\right)}{y^2}\frac{dy}{dx}+\frac{1}{y}{\varphi }^{{}^{\prime }}\left(x\right)=1$
(Product Rule)
$\frac{d}{dx}(\frac{1}{y}.\varphi (x\left)\right)=1$
$\frac{\varphi \left(x\right)}{y}=x+c$
$\varphi \left(1\right)=1;y\left(1\right)=1$
$1=1+c\Rightarrow c=0$
$\varphi \left(x\right)=xy$
$\Rightarrow y=\frac{\varphi \left(x\right)}{x}$
$\frac{1}{81}y\left(4\right)=\frac{1}{81}\frac{\varphi \left(4\right)}{4}=\frac{1}{81}\times \frac{1296}{4}=4$