Question

xd^2y/dx^2+dy/dx+xy=0​

Answer 1

Given to solve,

$\quad$

$x\cdot\dfrac {d^2y}{dx^2}+\dfrac{dy}{dx}+xy=0$

$\quad$

But we see that,

$\quad$

$x\cdot\dfrac {d^2y}{dx^2}+1\cdot\dfrac{dy}{dx}+xy=0\\\\\\x\cdot\dfrac {d}{dx}\left (\dfrac {dy}{dx}\right)+\dfrac {dx}{dx}\cdot\dfrac{dy}{dx}+xy=0$

$\quad$

That is,

$\quad$

$\dfrac {d}{dx}\left (x\cdot\dfrac{dy}{dx}\right)=-xy\\\\\\x\cdot\dfrac{dy}{dx}=\displaystyle-\int xy\ dx\\\\\\x\cdot\dfrac{dy}{dx}=\dfrac {1}{2}\int x^2\left (\dfrac {dy}{dx}\right)\ dx-x\cdot\dfrac {dy}{dx}\\\\\\4x\cdot\dfrac{dy}{dx}=\int x^2\left (\dfrac {dy}{dx}\right)\ dx\\\\\\\dfrac {d}{dx}\left (4x\cdot\dfrac{dy}{dx}\right)=x^2\cdot\dfrac {dy}{dx}\\\\\\4\cdot\dfrac{dy}{dx}+4x\cdot\dfrac{d^2y}{dx^2}=x^2\cdot\dfrac {dy}{dx}$

$4x\cdot\dfrac{d^2y}{dx^2}=(x^2-4)\cdot\dfrac {dy}{dx}$

$\quad$

Here, let,

$\quad$

$u=\dfrac {dy}{dx}$

$\quad$

Then the final equation becomes,

$\quad$

$4x\cdot\dfrac{du}{dx}=(x^2-4)u\\\\\\\dfrac {1}{u}\ du=\dfrac {x^2-4}{4x}\ dx$

$\quad$

Now,

$\quad$

$\displaystyle\int\dfrac {1}{u}\ du=\dfrac {1}{4}\int\dfrac {x^2-4}{x}\ dx\\\\\\\ln|u|+\ln|c_1|=\dfrac {1}{4}\int\left (x-\dfrac {4}{x}\right)\ dx\\\\\\\ln|u|+\ln|c_1|=\dfrac {x^2}{8}-\ln|x|+\ln|c_2|\\\\\\\ln\left|\dfrac {dy}{dx}\cdot c_1\right|=\dfrac {x^2}{8}-\ln\left|\dfrac {x}{c_2}\right|\\\\\\\dfrac {dy}{dx}\cdot c_1=e^{\frac {x^2}{8}-\ln\left|\frac {x}{c_2}\right|}\\\\\\\dfrac {dy}{dx}=\dfrac {e^{\frac {x^2}{8}}\cdot c_2}{c_1x}$

$\quad$

Let $\dfrac {c_1}{c_2}=C.$ Then,

$\quad$

$\dfrac {dy}{dx}=C\cdot\dfrac {e^{\frac {x^2}{8}}}{x}\\\\\\y=C\displaystyle\int\dfrac {e^{\frac {x^2}{8}}}{x}\ dx$

$\quad$

Let,

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$v=\dfrac {x^2}{8}\quad\iff\quad x^2=8v\\\\\\\dfrac {dv}{dx}=\dfrac {x}{4}\quad\iff\quad dx=\dfrac {4}{x}\ dv$

$\quad$

Then,

$\quad$

$y=C\displaystyle\int\dfrac {e^v}{x}\cdot\dfrac {4}{x}\ dv\\\\\\y=4C\displaystyle\int\dfrac {e^v}{x^2}\ dv\\\\\\y=\dfrac {C}{2}\displaystyle\int\dfrac {e^v}{v}\ dv\\\\\\y=\dfrac {C}{2}\displaystyle\int\dfrac {e^v}{v}\ dv\\\\\\\underline {\underline {y=\dfrac {C\cdot\text{Ei}\left (\frac {x^2}{8}\right)}{2}+c}}$

$\quad$

for some constants $c_1,\ c_2,\ c$ where $c_1,\ c_2\ \textgreater\ 1.$