# Differential equation - has y^2 [Solved!]

**Aage** 20 Oct 2017, 08:54

### My question

I am seraching a solution of:

c1*y^2+c2+dy/dx = 0

c1 and c2 are constants

### Relevant page

1. Solving Differential Equations

### What I've done so far

Different math books, have not been able to find a solution.

X

I am seraching a solution of:
c1*y^2+c2+dy/dx = 0
c1 and c2 are constants

Relevant page
<a href="https://www.intmath.com/differential-equations/1-solving-des.php">1. Solving Differential Equations</a>
What I've done so far
Different math books, have not been able to find a solution.

## Re: Differential equation - has y^2

**Murray** 20 Oct 2017, 23:03

@Aage: This is actually a **non-**linear differential equation, which is beyond the scope of the IntMath Forum. This PDF has a similar example which you may find useful: An Example of a Non-linear Differential Equation

X

@Aage: This is actually a <b>non-</b>linear differential equation, which is beyond the scope of the IntMath Forum. This PDF has a similar example which you may find useful: <a href="https://math.dartmouth.edu/archive/m23s06/public_html/handouts/nonlinear_example.pdf">An Example of a Non-linear Differential Equation</a>

## Re: Differential equation - has y^2

**stephenB** 14 Oct 2019, 07:06

You should not dismiss the problem as nonlinear before trying the frobenius method, revealing the solution y = (c2/c1)/x. This means there exists some transformation to a linear system, provided the singular point is removed.

Alternatively, you can solve this problem simply by deducing the proper transform: z=1/y.

The resulting ode factors:

(1/z^2)(1-c2/c1*dz/dx) =0.

So for y not zero, the solution (above) exists linearly.

Thanks for your website.

X

You should not dismiss the problem as nonlinear before trying the frobenius method, revealing the solution y = (c2/c1)/x. This means there exists some transformation to a linear system, provided the singular point is removed.
Alternatively, you can solve this problem simply by deducing the proper transform: z=1/y.
The resulting ode factors:
(1/z^2)(1-c2/c1*dz/dx) =0.
So for y not zero, the solution (above) exists linearly.
Thanks for your website.

## Re: Differential equation - has y^2

**Murray** 23 Dec 2019, 01:25

@stephenB: Thanks for your input.

X

@stephenB: Thanks for your input.

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