Solve the $1$st order DE $x^2y'+3xy=1$.

Question

I try dividing $x^2$ for both side. Then, I don't know what should I do next.

Answer 1

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 $$ y=\frac{c}{x^3}+\frac{1}{2x} $$

Answer 2

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 $$y=y_h+y_p=\frac{c}{x^3}+\frac{k}{x}$$

Answer 3

$${ x }^{ 2 }\frac { dy }{ dx } +3xy=1\\ \\ { x }^{ 2 }\frac { dy }{ dx } +3xy=0\\ \frac { dy }{ dx } =-3\frac { y }{ x } \\ \int { \frac { dy }{ y } } =-3\int { \frac { 1 }{ x } dx } \\ \ln { \left| y \right| =-3\ln { \left| cx \right| } } \\ y=\frac { C }{ { x }^{ 3 } } \\ y=\frac { C\left( x \right) }{ { x }^{ 3 } } \\ \frac { dy }{ dx } =\frac { { C }^{ \prime }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right) }{ { x }^{ 6 } } \\ { x }^{ 2 }\frac { { C }^{ \prime }\left( x \right) { x }^{ 3 }-3{ x }^{ 2 }C\left( x \right) }{ { x }^{ 6 } } +3x\frac { C }{ { x }^{ 3 } } =1\\ { C }^{ \prime }\left( x \right) =x\\ { C }\left( x \right) =\frac { { x }^{ 2 } }{ 2 } +C_{ 1 }\\ y=\frac { \frac { { x }^{ 2 } }{ 2 } +C_{ 1 } }{ { x }^{ 3 } } =\frac { 1 }{ 2{ x } } +\frac { C_{ 1 } }{ { x }^{ 3 } } \\ \\ $$

Answer 4

$$x^2y'+3xy=1$$ 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 $$A=\frac{1}{2}$$

$$y=y_c+y_p=\frac{C_1}{x^3}+\frac{1}{2x}$$

Answer 5

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 $y$.