Question

xdy/dx + y = y² logx​

Answer 1

$\displaystyle\large\boxed {\sf {y=\dfrac {1}{1+\log x}}}$

Given,

$\displaystyle\longrightarrow\sf{x\ \dfrac {dy}{dx}+y=y^2\log x}$

Dividing each term by x, we get,

$\displaystyle\longrightarrow\sf{\dfrac {dy}{dx}+\dfrac {y}{x}=\dfrac {y^2}{x}\log x}$

But we have to make RHS in such a way that it should contain a function in x only, to get a linear differential equation. So, dividing each term in the equation by $\displaystyle\sf {y^2,}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{y^2}\cdot\dfrac {dy}{dx}+\dfrac {1}{xy}=\dfrac {1}{x}\log x\quad\quad\dots (1)}$

Let,

$\displaystyle\longrightarrow\sf{du=\dfrac {1}{y^2}\ dy}$

$\displaystyle\longrightarrow\sf{u=\int\dfrac {1}{y^2}\ dy}$

$\displaystyle\longrightarrow\sf{u=\dfrac {y^{-1}}{-1}}$

[no integral constant is considered or it is taken as zero.]

$\displaystyle\longrightarrow\sf{-u=\dfrac {1}{y}}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{\dfrac {du}{dx}-\dfrac {u}{x}=\dfrac {1}{x}\log x}$

Now it is being a linear differential equation in the form $\displaystyle\sf{\dfrac {du}{dx}+Pu=Q}$ where $\displaystyle\sf {P=-\dfrac {1}{x}}$ and $\displaystyle\sf {Q=\dfrac {1}{x}\log x.}$

Then, multiply each term by $\displaystyle\sf {f(x).}$

$\displaystyle\longrightarrow\sf{f(x)\cdot\dfrac {du}{dx}+P\cdot f(x)\cdot u=Q\cdot f(x)\quad\quad\dots (2)}$

Choose $\displaystyle\sf {f(x)}$ such that $\displaystyle\sf {Q\cdot f(x)=\dfrac {d}{dx}[u\cdot f(x)],}$ i.e.,

$\displaystyle\longrightarrow\sf{f(x)\cdot\dfrac {du}{dx}+P\cdot f(x)\cdot u=\dfrac {d}{dx}[u\cdot f(x)]}$

$\displaystyle\longrightarrow\sf{f(x)\cdot\dfrac {du}{dx}+P\cdot f(x)\cdot u=f(x)\cdot\dfrac {du}{dx}+u\cdot\dfrac {d}{dx}[f(x)]}$

$\displaystyle\longrightarrow\sf{P\cdot f(x)=\dfrac {d}{dx}[f(x)]\quad\quad\dots (i)}$

Then in (2),

$\displaystyle\longrightarrow\sf{f(x)\cdot\dfrac {du}{dx}+\dfrac {d}{dx}[f(x)]\cdot u=Q\cdot f(x)}$

$\displaystyle\longrightarrow\sf{\dfrac {d}{dx}[u\cdot f(x)]=Q\cdot f(x)}$

$\displaystyle\longrightarrow\sf{u\cdot f(x)=\int Q\cdot f(x)\ dx}$

$\displaystyle\longrightarrow\sf{u=\dfrac {1}{f(x)}\int Q\cdot f(x)\ dx\quad\quad\dots (3)}$

From (i),

$\displaystyle\longrightarrow\sf{P=\dfrac {\dfrac {d}{dx}[f(x)]}{f(x)}}$

On integrating each term wrt x, we get,

$\displaystyle\longrightarrow\sf{\int P\ dx=\log (f(x))}$

$\displaystyle\longrightarrow\sf{f(x)=e^{\int P\ dx}}$

Thus (3) becomes,

$\displaystyle\longrightarrow\sf{u=e^{-\int P\ dx}\int\left (Q\cdot e^{\int P\ dx}\right)\ dx}$

Undo $\displaystyle\sf {u=-\dfrac {1}{y},\ P=-\dfrac {1}{x}}$ and $\displaystyle\sf {Q=\dfrac {1}{x}\log x.}$ Then,

$\displaystyle\longrightarrow\sf{-\dfrac {1}{y}=e^{\int\frac {1}{x}\ dx}\int\left (\dfrac {1}{x}\log x\cdot e^{-\int\frac {1}{x}\ dx}\right)\ dx}$

$\displaystyle\longrightarrow\sf{-\dfrac {1}{y}=e^{\log x}\int\left (\dfrac {1}{x}\log x\cdot e^{-\log x}\right)\ dx}$

$\displaystyle\longrightarrow\sf{-\dfrac {1}{y}=x\int\left (\dfrac {1}{x}\log x\cdot\dfrac {1}{x}\right)\ dx}$

$\displaystyle\longrightarrow\sf{-\dfrac {1}{y}=x\int\left (\dfrac {1}{x^2}\log x\right)\ dx\quad\quad\dots (4)}$

Well, by product rule,

$\displaystyle\longrightarrow\sf{\int\dfrac {1}{x^2}\log x\ dx=-\dfrac {1}{x}\log x+\int\dfrac {1}{x}\cdot\dfrac {1}{x}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {1}{x^2}\log x\ dx=-\dfrac {1}{x}\log x+\int\dfrac {1}{x^2}\ dx}$

$\displaystyle\longrightarrow\sf{\int\dfrac {1}{x^2}\log x\ dx=-\dfrac {1}{x}\log x-\dfrac {1}{x}}$

$\displaystyle\longrightarrow\sf{\int\dfrac {1}{x^2}\log x\ dx=-\dfrac {1}{x}[1+\log x]}$

Then (4) becomes,

$\displaystyle\longrightarrow\sf{-\dfrac {1}{y}=-x\cdot\dfrac {1+\log x}{x}}$

$\displaystyle\longrightarrow\sf {\underline {\underline {y=\dfrac {1}{1+\log x}}}}$