I $\heartsuit$ Substitution!

Let's start with the given differential equation:

$\frac{dy}{dx}+\frac{y}{x}=\ln(x)y^{2}$

Now, divide both sides by $-y^{-2}$ to get:

$-y^{-2}\frac{dy}{dx}-\frac{1}{xy}=-ln(x)$

Let $w=y^{-1}$ . This means that $\frac{dw}{dx}=-y^{-2}\frac{dy}{dx}$ . Substituting this in gets us:

$\frac{dw}{dx}-\frac{w}{x}=-ln(x)$

Multiply both sides by $x$ to get:

$x\frac{dw}{dx}-w=-xln(x)$

Now comes the most crucial step - Let $w=ux$ for some parameterization $u$ . This means that $\frac{dw}{dx}=u+x\frac{du}{dx}$ . Now substituting this in gets us:

$ux+x^{2}\frac{du}{dx}-ux=-xln(x)$ $\frac{du}{dx}=\frac{-ln(x)}{x}$

Now this looks like something we know how to deal with!

$\int { du } =\int { \frac { -\ln { \left( x \right) } }{ x } dx }$

Let $v=ln(x)$ . Then $dv=\frac{1}{x}dx$ , and so we have:

$\int { du } =\int { -v \quad dv }$

$u=\frac{-1}{2}v^{2}+k$

$u=\frac{-1}{2}ln^{2}(x)+k$

From the above information we know that $u=\frac{w}{x}=\frac{1}{xy}$ , so:

$\frac{1}{xy}= \frac{-1}{2}ln^{2}(x)+k$

And finally:

$y=\frac{1}{kx-\frac{1}{2}xln^{2}(x)}$

In order to get the equation in the form that the question asked for, multiply the RHS by $\frac{2}{2}$ and pull out the $x$ in the denominator in order to get:

$y=f(x)=\frac{2}{x(2k-ln^{2}(x))}$

Now for the Initial Value piece of the problem. We know that $f(1)=1$ , and that

$f(1)=\frac{2}{2k}$

so $2k=2$ , and we can FINALLY write $f(x)$ as

$f(x)=\frac{2}{x(2-ln^{2}(x))}$

and so $a+b+c+d=2+1+2+2=\boxed{7}$ .

$\dfrac{dy}{dx} + \dfrac{y}{x} = \ln x\cdot y^{2}$
Dividing by $y^{2}$
$\dfrac{1}{y^{2}} \dfrac{dy}{dx} + \dfrac{1}{xy} = \ln x$
$\dfrac{1}{y} = t \rightarrow \dfrac{1}{y^{2}} \dfrac{dy}{dx} = \dfrac{-dt}{dx}$
The differential transforms into,
$\dfrac{dt}{dx} - \dfrac{t}{x} = - \ln x$
This is a linear differential equation of the form $\dfrac{dt}{dx} + Pt = Q$
Integrating factor = $e^{ \int P dx} = e^{ \int \dfrac{-dx}{x}} = \dfrac{1}{x}$
Multiplying by $\dfrac{1}{x}$ on both sides,

$\dfrac{dt}{xdx} - \dfrac{t}{x^{2}} = \dfrac{- \ln x}{x}$
$\dfrac{d\left(\dfrac{t}{x}\right)}{dx} = \dfrac{-\ln x}{x}$
$\displaystyle \int d\left( \dfrac{t}{x} \right)= \displaystyle \int \dfrac{-\ln x}{x}dx + \dfrac{c}{2}$
Integrating both sides,
$\dfrac{2t}{x} = c - \ln^{2} x$
$\dfrac{2}{xy} = c - \ln^{2} x$
Since, $f(1) = 1$
$\dfrac{2}{1\cdot1} = c - 0 \rightarrow c = 2$
$\therefore y = \dfrac{2}{x(2-\ln^{2} x)}$
Comparing, a = 2 , b = 1, c = 2, d = 2
$a + b + c + d= 2 + 2 + 2 + 1 = 7$

First, rewrite the original equation as, $x \frac{dy}{dx} + y = xy^2 \ln(x)$ in which can be written as, $\frac{d(xy)}{dx} = xy^2 \ln(x)$ Define $u = xy$ , then we have a separable differential equation, $\frac{du}{dx} = \frac{u^2}{x}\ln(x)$ Since $\int\:\frac{\ln(x)}{x}\:dx = \frac{1}{2}\ln^2(x)+C$ , then we have $u(x) = -\frac{1}{\frac{1}{2}\ln^2(x)+C}$ or, in original variable, $y(x) = -\frac{1}{x(\frac{1}{2}\ln^2(x)+C)}$ The condition $y(1)=1$ imply $C=-1$ . Hence, we have $y(x) = \frac{2}{x(2-\ln^2(x))}$

So, $a+b+c+d = 7$ .