Solving Differential Equations of the Form y'+Cy^2=D

[gamesguru]

I'm curious about the differential equation which takes the general form of,
$$y'+Cy^2=D$$.
Where C and D are constants. According to mathematica, the answer is:
$$\sqrt{\frac{D}{C}} \tanh{(x \sqrt{CD})}$$.
But I'd like to know how this is done by hand, I was able to do this with separation of variables and finding x, then solving for y, but it was a large waste of time and I'm curious if there is a general way to solve this just as linear ones can be solved.

 

[lurflurf]

no that is why linear equations are nice they are much easier to solve and can always (at least in principle) be solved. It is a simple integration.

 

[foxjwill]

Actually, yes, it can. http://mathworld.wolfram.com/RiccatiDifferentialEquation.html . Look at equation (4).

 

[HallsofIvy]

I don't know why you would use the Riccatti equation. This is a simple, separable, first order differential equation. It can be integrated directly. I don't know exactly what gamesguru did or why he considers it a "large waste of time" but it is not difficult. It separates into
[tex]\frac{dy}{Cy^2+ D}= dx[/itex]
multiplying on both sides by D, we get
$$\frac{dy}{\frac{C}{D}y^2+ 1}= Ddx$$
Now, what we do depends upon the signs of C and D. If C and D have opposite signs so that C/D is negative, write the equation as
$$\frac{dy}{1- \left|\frac{C}{D}}\right|y^2}= Ddx$$
and let $u= \sqrt{|C/D|}$. Then the equation becomes
$$\sqrt{|C/D|}\frac{du}{u^2+ 1}= Ddt$$
and the solution is exactly what gamesguru said.

If C and D have the same sign, let u= $\sqrt{C/D}y$ so that the equation becomes
$$\sqrt{\frac{D}{C}}\frac{du}{u^2+ 1}= Ddt$$
and integrating gives the same thing with "tan" rather than "tanh".

 

[exk]

Shouldn't it be:
$$\frac{dy}{D-Cy^2}= dx$$
?
The answer changes, but the method is the same.

 

[foxjwill]

I don't know why you would use the Riccatti equation.

Yes, but he wanted to know if there was a way to do it without separation of variables.