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Solve the equation:

$xdy=(x^5+x^3y^2+y)dx$

My attempt at a solution: I've tried to check if this was an exact differential equation or if I could reduce it to one, so

$xdy=(x^5+x^3y^2+y)dx \iff xdy-(x^5+x^3y^2+y)dx=0$. If I call $M(x,y)=-(x^5+x^3y^2+y)$ and $N(x,y)=x$, the equation is exact if and only if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x} \iff -(2x^3y+1)=1$, which is clearly false. I've tried to reduce this to an exact differential equation:

I suppose there exists a function $u$ that depends only on $x$ or on $y$ such that

$(uM)dx+(uN)dy=0$

Case 1: if $u$ is $u(x)$, then $u$ must satisfy

$\dfrac{u_x}{u}=\dfrac{M_y-N_x}{N} \iff \dfrac{u_x}{u}=\dfrac{-1-2yx^3-1}{x}$ But the member of the right doesn't depende only on $x$, so I can't find $u(x)$ that satisfies the required conditions.

Case 2: I suppose there exists $u$ that satisfies $(uM)dx+(uN)dy=0$ but that only dependes on $y$. So $\dfrac{u_y}{u}=\dfrac{N_x-M_y}{M}$. Here I had the same problem, the member of the right doesn't depend only on $y$.

What other method could I apply to solve the ODE? Should I propose a function $u$ that depends on both $x$ and $y$? I don't know how to find $u$ if this is the case.

Here's the solution with Amzoti's answer: Suppose $y=vx \implies y'=v'x+v$, replacing this into the ODE, we get $xdy=(x^5+x^3y^2+y)dx \iff x\dfrac{dy}{dx}=x^5+x^3y^2+y \iff x(v'x+v)=x^5+v^2x^5+vx \iff v'x+v=x^4+v^2x^4+v \iff v'x=x^4(1+v^2) \iff v'=x^3(1+v^2)$. But $v'=x^3(1+v^2)$ is a separable equation, dividing by $1+v^2$ and multiplying by $dx$, we get $\dfrac{dv}{1+v^2}=x^3dx$, integrating we get that

$arctan(v)=\dfrac{x^4}{4}+c, c \in \mathbb R$. Then, the implicit solution of the original ODE is $arctan(\dfrac{y}{x})==\dfrac{x^4}{4}+c , c \in \mathbb R$

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Hint: Let:

$$y = v x \rightarrow y' = v + x v'$$

Substitute this into the ODE and solve. It will reduce it to a separable equation and you can use integration of both side.

It may also be possible to get this in Riccati form.

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  • $\begingroup$ In memory of J.F.Ricatti $(1676-1754)$ :+) $\endgroup$ – mrs Nov 25 '13 at 2:37

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