# Solve differential equation $xyy'=x^4+y^4$

How to find general solution to this differential equation (if it exists): $$xyy'=x^4+y^4 ?$$ I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $$x$$ and $$y$$ are symetric in equation: $$y'=\dfrac{x^4+y^4}{xy}$$ so I expect (maybe) some kind of symetric function for y(x).

Thanks for any help.

• Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck – poetasis Jan 12 at 17:19
• @poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$? – Thom Jan 12 at 17:23
• I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more. – poetasis Jan 12 at 17:25
• Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4)\,dx-xy\,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants. – Adrian Keister Jan 12 at 17:45
• The definition of a new dependent variable $z := \frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$. – Christoph Jan 12 at 17:49

Observing that $$y y' = \left(\frac{1}{2}y^2\right)'$$ we define the new dependent variable $$z := \frac{1}{2} y^2$$, $$y^4 = 4 z^2$$. We then obtain a Riccati equation for $$z$$: $$z' = x^3 + \frac{4}{x} z^2$$. This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing $$$$z = - \frac{x}{4} \frac{u'}{u}, \quad z^2 = \frac{x^2}{16} \frac{(u')^2}{u^2}, \quad z' = - \frac{1}{4} \frac{u'}{u} - \frac{x}{4} \frac{u'' u - (u')^2}{u^2},$$$$ which yields $$x^2 u'' + x u' + 4 x^4 u = 0$$. With the definition of a new independent variable $$$$\xi := x^2, \quad \frac{d}{dx} = 2 \xi^{1/2} \frac{d}{d\xi}, \quad \frac{d^2}{dx^2} = 2 \frac{d}{d\xi} + 4 \xi \frac{d^2}{d\xi^2},$$$$ we obtain the Bessel differential equation $$$$\xi^2 \frac{d^2 u}{d \xi^2} + \xi \frac{d u}{d\xi} + \xi^2 u = 0,$$$$ with fundamental solutions $$J_0(\xi)$$, $$Y_0(\xi)$$ (zeroth-order Bessel functions).
The text passage contains a mathematical question in the form of a differential equation and provides a detailed solution approach, including transformations and reductions to more manageable forms of the equation.
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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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# Solve differential equation $xyy'=x^4+y^4$

How to find general solution to this differential equation (if it exists): $$xyy'=x^4+y^4 ?$$ I do not know how to even approach it since I never dealt with nonlinear equations. Only thing that I notice is that $$x$$ and $$y$$ are symetric in equation: $$y'=\dfrac{x^4+y^4}{xy}$$ so I expect (maybe) some kind of symetric function for y(x).

Thanks for any help.

• Try an indefinite integral on both sides of the equation. It still won't be easy but there are tools to help. Good luck – poetasis Jan 12 at 17:19
• @poetasis In the link you sent there is step where $y$ is taken out of integral, why is that posible if $y$ itself is function of $x$? – Thom Jan 12 at 17:23
• I didn't do the calculation; the tool did and it's been 40 years since I got my degree so I'm rusty on this stuff. I wish I could help you more. – poetasis Jan 12 at 17:25
• Couple of comments: 1. It isn't actually symmetric, as you can see by rearranging into differential form: $(x^4+y^4)\,dx-xy\,dy=0.$ 2. The solution, if it exists, must have positive slope in the first and third quadrants, and negative slope in the second and fourth quadrants. – Adrian Keister Jan 12 at 17:45
• The definition of a new dependent variable $z := \frac{1}{2} y^2$ yields a Riccati equation for $z$: $z' = x^3 + \frac{4}{x} z^2$. – Christoph Jan 12 at 17:49

Observing that $$y y' = \left(\frac{1}{2}y^2\right)'$$ we define the new dependent variable $$z := \frac{1}{2} y^2$$, $$y^4 = 4 z^2$$. We then obtain a Riccati equation for $$z$$: $$z' = x^3 + \frac{4}{x} z^2$$. This Riccati equation (like any Riccati equation) can be reduced to a second-order linear ordinary differential equation by writing $$$$z = - \frac{x}{4} \frac{u'}{u}, \quad z^2 = \frac{x^2}{16} \frac{(u')^2}{u^2}, \quad z' = - \frac{1}{4} \frac{u'}{u} - \frac{x}{4} \frac{u'' u - (u')^2}{u^2},$$$$ which yields $$x^2 u'' + x u' + 4 x^4 u = 0$$. With the definition of a new independent variable $$$$\xi := x^2, \quad \frac{d}{dx} = 2 \xi^{1/2} \frac{d}{d\xi}, \quad \frac{d^2}{dx^2} = 2 \frac{d}{d\xi} + 4 \xi \frac{d^2}{d\xi^2},$$$$ we obtain the Bessel differential equation $$$$\xi^2 \frac{d^2 u}{d \xi^2} + \xi \frac{d u}{d\xi} + \xi^2 u = 0,$$$$ with fundamental solutions $$J_0(\xi)$$, $$Y_0(\xi)$$ (zeroth-order Bessel functions).
user
The text passage contains a mathematical question in the form of a differential equation and provides a detailed solution approach, including transformations and reductions to more manageable forms of the equation.

TYPE: SINGLE_QUESTION_WITH_ANSWER