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 sum of digits of value of $y\left(4\right)$ is equal to

Answer 1

The given equation can be written as
$\frac{dy}{dx}-y\frac{{\varphi }^{\prime }\left(x\right)}{\varphi \left(x\right)}=\frac{-y^2}{\varphi \left(x\right)}$
$\Rightarrow -\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}\frac{{\varphi }^{\prime }\left(x\right)}{\varphi \left(x\right)}=\frac{1}{\varphi \left(x\right)}\left(i\right)$
Put $\frac{1}{y}=u$ so that $-\frac{1}{y^2}\frac{dy}{dx}=\frac{du}{dx}$. Now (i) becomes $\frac{du}{dx}+u\frac{{\varphi }^{\prime }\left(x\right)}{\varphi \left(x\right)}=\frac{1}{\varphi \left(x\right)}$
This is a linear equation in u so its integrating factor is
$e^{\int \frac{{\varphi }^{\prime }\left(x\right)}{o\left(x\right)}dx}=e^{\log \varphi \left(x\right)}=\varphi \left(x\right)$
Multiplying both sides by the integrating factor, we have
$\frac{d}{dx}\left[u\varphi \right(x\left)\right]=1\Rightarrow u\varphi \left(x\right)=x+constant\Rightarrow y=\frac{\varphi \left(x\right)}{x+C}$
$\Rightarrow y\left(1\right)=\frac{1}{1+C}\Rightarrow C=0$. Thus $y=\frac{\varphi \left(x\right)}{x}$
Hence $y\left(4\right)=\frac{1296}{4}=324$