Solve the $1$st order DE $x^2y'+3xy=1$. I try dividing $x^2$ for both side. Then, I don't know what should I do next.
 A: Again substitute $y(x)=z(x)/x$ solution you find $x z'(x)=1-2 z(x)$ and then (by separating the variable) 
$$
-\frac{1}{2}\log(1-2z)=c+\log(x)
$$
or
$$
z=\frac{1}{2}+\frac{c}{x^2}
$$
or
$$
y=\frac{c}{x^3}+\frac{1}{2x}
$$
A: Rewrite 
$$y'=-3\frac{y}{x}+\frac{1}{x^2}$$
Solving the homogenous equation is pretty eays (e.g. using method of separation) $$y_h=c_1x^{-3}$$
Then you guess (is faster for easy ODEs) a particular solution of the form $y_p=\frac{k}{x}$. Plug this into the equation and find k. 
Note: You can also get the particular solution using the general formula for a linear ODE or variation of constants.
As always you general solution is 
$$y=y_h+y_p=\frac{c}{x^3}+\frac{k}{x}$$
A: $${ x }^{ 2 }\frac { dy }{ dx } +3xy=1\\ \\ { x }^{ 2 }\frac { dy }{ dx } +3xy=0\\ \frac { dy }{ dx } =-3\frac { y }{ x } \\ \int { \frac { dy }{ y }  } =-3\int { \frac { 1 }{ x } dx } \\ \ln { \left| y \right| =-3\ln { \left| cx \right|  }  } \\ y=\frac { C }{ { x }^{ 3 } } \\ y=\frac { C\left( x \right)  }{ { x }^{ 3 } } \\ \frac { dy }{ dx } =\frac { { C }^{ \prime  }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right)  }{ { x }^{ 6 } } \\ { x }^{ 2 }\frac { { C }^{ \prime  }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right)  }{ { x }^{ 6 } } +3x\frac { C }{ { x }^{ 3 } } =1\\ { C }^{ \prime  }\left( x \right) =x\\ { C }\left( x \right) =\frac { { x }^{ 2 } }{ 2 } +C_{ 1 }\\ y=\frac { \frac { { x }^{ 2 } }{ 2 } +C_{ 1 } }{ { x }^{ 3 } } =\frac { 1 }{ 2{ x } } +\frac { C_{ 1 } }{ { x }^{ 3 } } \\ \\ $$
A: $$x^2y'+3xy=1$$
divide by $x$
$$xy'+3y=\frac{1}{x}$$
we can solve this O.D.E by Euler-Cauchy Method
1- to find the complementary solution
$$xy'+3y=0$$
assume 
$$y_c=x^m$$
$$y'=mx^{m-1}$$
substitute it to get
$$m=-3$$
hence
$$y_c=C_1x^{-3}=\frac{C_1}{x^3}$$ 
2- to find the particular solution 
         $$y_p=\frac{A}{x}$$
         $$y'_p=-\frac{A}{x^2}$$
   substitute it to get
    $$A=\frac{1}{2}$$
$$y=y_c+y_p=\frac{C_1}{x^3}+\frac{1}{2x}$$
A: When you have a first order non-linear and/or non-autonomous differential equation you could always try testing if it is an exact differential equation
$$
p(x,y) + q(x,y) \frac{dy}{dx} = 0,
$$
where
$$
p(x,y) = 3xy - 1,
$$
$$
q(x,y) = x^2.
$$
In this case $\frac{\partial p}{\partial y} \neq \frac{\partial q}{\partial x}$, however if you multiply this equation by some other function $\mu(x,y)$ the new $p(x,y)$ and $q(x,y)$ might satisfy this relation.
Since all terms in the existing differential equation only contain powers of $x$ and $y$, a good guess would be
$$
\mu(x,y) = x^a y^b.
$$
The values for $a$ and $b$ can be found by equating
$$
\frac{\partial \mu(x,y) p(x,y)}{\partial y} = \frac{\partial \mu(x,y) q(x,y)}{\partial x},
$$
which is equal to $\frac{\partial}{\partial x} \frac{\partial}{\partial y} \Psi(x,y)$, such that differential equation is equivalent to
$$
\frac{\partial}{\partial x} \Psi(x,y) + \frac{\partial}{\partial y} \Psi(x,y) \frac{dy}{dx} = \frac{d}{dx} \Psi(x,y) = 0,
$$
thus $\Psi(x,y)$ has to be equal to a constant, represented by $c$.
The solution to the differential equation can then be found by integrating to find $\Psi(x,y)$
$$
\Psi(x,y) = \int{\mu(x,y) p(x,y) dx} + f(y),
$$
$$
\Psi(x,y) = \int{\mu(x,y) q(x,y) dy} + g(x).
$$
Both integrals will contain the same terms, which both depend on $x$ and $y$, such that $g(x)$ will be equal to the terms of the first integral, which only depends on $x$ and $f(y)$ will be equal to the terms of the second integral, which only depends on $y$.
