Solve the equation:


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


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 $arctan(\dfrac{y}{x})==\dfrac{x^4}{4}+c , c \in \mathbb R$


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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.