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Solve the differential equation $$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$

My try:

we can write the equation as:

$$\frac{dy}{dx}=\frac{1}{y^3}\frac{\left(1+2y^4\right)}{1+\frac{4x}{y^4}}$$

Multiplying both sides with $\frac{1}{y^5}$ we get:

$$\frac{1}{y^5}\frac{dy}{dx}=\frac{1}{y^8}\frac{y^4(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}=\frac{1}{y^4}\frac{(2+\frac{1}{y^4})}{1+\frac{4x}{y^4}}$$

Now letting $$\frac{1}{y^4}=t$$ we get

$$\frac{-1}{4}\frac{dt}{dx}=\frac{t^2+2t}{4tx+1}$$

Any way further to convert in to variable separable?

Umesh shankar
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2 Answers2

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This is an "almost" exact equation. Write it in $A(x,y)\text dx + B(x,y)\text dy=0$ form

$$(-y-2y^5)\text dx + (4x+y^4)\text dy = 0$$

Then, look for a function $\mu(y)$ such that

$$ (-y-2y^5)\mu(y)\text dx + (4x+y^4)\mu(y)\text dy = 0 $$

Let $\mu(y) = y^{-5}$. Then, the equation becomes

$$ (-y^{-4}-2)\text dx + (4xy^{-5}+y^{-1})\text dy = 0 \\ \text d\left[ -xy^{-4} - 2x + \log y \right] = 0$$

Integrating, we get

$$ -xy^{-4} - 2x + \log y = C $$

Solving for $x$ gives us

$$ x = \frac{y^4}{2y^4+1}\log(cy) $$

Solving for $y$ in terms of the function $W(z)$ such that $W(z)\exp(W(z)) = W(z\exp(z)) = z$,

$$y = \left( \frac{4x}{W(kx\exp(-8x))} \right)^\frac{1}{4}$$

Alex Jones
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$$\frac{dy}{dx}=\frac{y+2y^5}{4x+y^4}$$

$$(y+2y^5)\frac{dx}{dy}-4x=y^4$$ Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is : $$x(y)=\frac{y^4}{2y^4+1}\ln(c\:y)$$ $y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function. $$y(x)=\left(\frac{4x}{\text{W}(C\:xe^{-8x})}\right)^{1/4}$$ $C$ is an arbitrary constant, to be determined according to some boundary condition.

JJacquelin
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