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How to solve $$\frac{dy}{dx}-1=\frac{x^2}{y}$$

I tried letting $y^2=u$ which gives $$y\frac{dy}{dx}=\frac{1}{2}\frac{du}{dx}$$

So the equation is now $$\frac{du}{dx}-2\sqrt{u}=2x^2$$ Any help?

MathematicianByMistake
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Ekaveera Gouribhatla
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  • @MorA. I'm not sure what that accomplishes. Perhaps you've mistaken this for a linear equation? – Eli Bartlett Sep 26 '22 at 19:15
  • @EliBartlett You're right, I made a mistake there – Mor A. Sep 26 '22 at 19:27
  • There are known exact solutions for similar problems, such as $yy'-y=Ax^2-9/(625A)$ and $yy'-y=\pm6x/25+Ax^2$ for arbitrary $A$. I have tried a couple transformations to arrive at a known equation to no avail. – Eli Bartlett Sep 29 '22 at 06:29
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    This equation is not analytically solvable yet in a closed form, as it is a type of Chini equation. You can get a series solution, but I think that's the best you can do. – worthlesspear Sep 29 '22 at 11:05
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    Even wolfram alpha can't solve this. Like previous comment said, it is a chini equation and only a few of them have known solutions. – DanielV Sep 29 '22 at 21:45

2 Answers2

0

Not an answer

As a possible aid to others in possibly guessing analytical solution curve shapes, integrands of

$$\frac{dy}{dx}-c=\frac{x^2}{y}$$

with initial slope values $ y^{'}_{i}=c $ at $x=0 $ are numerically computed and plotted. When $c=0,$ $y\to k\cdot x^{3/2}$

( fwiw with no possible further consequence its second order ode .. if I made no error in differentiating is)

$$ \frac{y y'''-y''}{y'y''}= (2 c-1). $$

enter image description here

Narasimham
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After reviewing comments:

use:

$x=\cos{\theta},\;y=\sin{\theta}$

Then:

$dy=\cos{\theta}d\theta,\;dx=-\sin{\theta}d\theta$. Substitute in : $\frac{dy}{dx}=1+\frac{x^2}{y}$

$\cos{\theta}d\theta=-\sin{\theta}d\theta+\frac{\cos^2{\theta}}{\sin{\theta}}\cdot(-\sin{\theta})d\theta$

So now everything is integrable.

$(\cos{\theta}+\sin{\theta}+\frac{1}{2}(1+\cos{2\theta}))d\theta=0$

$-\sin{\theta}+\cos{\theta}+\frac{1}{2}\theta+\frac{1}{4}\sin{2\theta}=c$

Substitute back.

$-y+x+\frac{1}{2}\arctan{\frac{y}{x}}+\frac{1}{2}xy=c$.

$c$ constant to be determined from initial condition.

ryaron
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  • No no you took the wrong question – Ekaveera Gouribhatla Sep 26 '22 at 10:54
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    Your first line should be $y, dy−(y+x^2)dx=0$ not $y, dy−(1+x^2)dx=0$ – EnEm Sep 26 '22 at 10:57
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    Now from your assumption $x=\cos \theta$ and $y=\sin \theta$, you can just say $y=\sqrt{1-x^2}$, you don't even have to do anything else. Basically you are limiting your domain to a very tiny set from $\mathbb{R}\times\mathbb{R}$, so the solution isn't that useful. – EnEm Sep 26 '22 at 12:56
  • I am defining it as a one parameter curve. – ryaron Sep 26 '22 at 15:43
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    The curve $x=\cos \theta$ and $y=\sin \theta$ is not the solution except there is some constant value $c$ such that for some range of $\theta$ equality $-\sin{\theta}+\cos{\theta}+\frac{1}{2}\theta+\frac{1}{4}\sin{2\theta}=c$ is satisfied. But this is the same as $\cos\theta+\sin\theta+\frac{1}{2}(1+\cos 2\theta)=0$ in some range which is impossible. – Ivan Kaznacheyeu Sep 26 '22 at 15:59