Differential equations - cannot solve one

[luuurey]

Greetings everyone.
Can you help me please with solving this differential equation?

$$\large \frac{dv}{dt} = a_g + \alpha v^n$$

where $$\Large a_g \alpha n$$ are constants. Artelnatively with specific n as n = 1, 2.

I have no idea what to do...
Thank you very much

 

[Millennial]

It seems quite simple at the first glance, as the equation is separable.

The first thing I'd do is to separate the two variables to get $\displaystyle \frac{dv}{a_g+\alpha v^n}=dt$. From there, you integrate both sides.

The "trick" here is that the left hand side doesn't seem integrable. In fact, it isn't very easy and pleasant to integrate for most n's, but it can be done. I can't find a general algorithm though.

 

[luuurey]

Okay. Let's say n = 1. Then it should be resolvable. And what if n = 2 ? For n = 0 it is trivial...

 

[Dick]

Okay. Let's say n = 1. Then it should be resolvable. And what if n = 2 ? For n = 0 it is trivial...

For n=1 a simple substitution will give you log. For n=2 a trig substitution will get you an arctan.

 

[luuurey]

Can you show me please ?

 

[SammyS]

Can you show me please ?

How does it give you arctan ?

What is $\displaystyle \ \ \int\frac{dx}{1+x^2}\ ?$

 

[HallsofIvy]

Have you never taken a Calucus class? If n= 0, The integral is just
$$\int \frac{dv}{a_g+ \alpha}= \frac{1}{a_g+ \alpha}\int dv$$

If n= 1 it is
$$\int\frac{dv}{a_g+ \alpha v}$$

If n= 2 it is
$$\int\frac{dv}{a_g+ \alpha v^2}$$

Can you integrate those?

 

[luuurey]

No, I haven't. I do not know how to substitute. The first one I can. The next two I cannot.

 

[SammyS]

Can you show me please ?

What is the derivative of arctan(x) ?

 

[JasonPhysicist]

No, I haven't. I do not know how to substitute. The first one I can. The next two I cannot.

For integrals of the type

$$\int \frac{dx}{a^2 +x^2}$$, try the following substitution :

$$x=a \tan \theta$$

With a bit of algebra, you'll be able to put your integral at this form.

 

[luuurey]

$$\int \frac{dx}{a^2 +x^2}$$
$$x=a \tan \theta$$
$$dx=\frac {a} {\cos^2 \theta} d \theta$$ - is it like this ?

so we get:
$$\frac {1}{a^2}\int \frac{\frac{a}{\cos^2}d \theta}{1 +\tan^2 \theta} = \frac {1}{a}\int \frac{d\theta}{\cos^2\theta +\sin^2 \theta} = \frac{\theta}{a} = \frac {\arctan \frac{x}{a}}{a}$$
Okay and what next ?

 

[JasonPhysicist]

$$\int \frac{dx}{a^2 +x^2}$$
$$x=a \tan \theta$$
$$dx=\frac {a} {\cos^2 \theta} d \theta$$ - is it like this ?

so we get:
$$\frac {1}{a^2}\int \frac{\frac{a}{\cos^2}d \theta}{1 +\tan^2 \theta} = \frac {1}{a}\int \frac{d\theta}{\cos^2\theta +\sin^2 \theta} = \frac{\theta}{a} = \frac {\arctan \frac{x}{a}}{a}$$
Okay and what next ?

Fine. Now, compare the two integrals and identify your "a".

 

[luuurey]

Well
$$a = \sqrt{a_g} \\ x = \sqrt{\alpha}v$$

So I got
$$\frac {\arctan({\sqrt{\frac{\alpha}{a_g}} v})}{\sqrt{a_g}} = t$$
didn't I?
So
$$v = \frac {\tan(\sqrt{a_g}t)}{\sqrt{\frac{\alpha}{a_g}}}$$
IS IT RIGHT ?

 

[JasonPhysicist]

Well
$$a = \sqrt{a_g} \\ x = \sqrt{\alpha}v$$

So I got
$$\frac {\arctan({\sqrt{\frac{\alpha}{a_g}} v})}{\sqrt{a_g}} = t$$
didn't I?
So
$$v = \frac {\tan(\sqrt{a_g}t)}{\sqrt{\frac{\alpha}{a_g}}}$$
IS IT RIGHT ?

Actually, do this:

$$\frac{1}{\alpha} \int \frac{dv}{\frac{a_g}{\alpha}+v^2}$$

Then, use the substitution to integrate.

 

[luuurey]

Actually, do this:

$$\frac{1}{\alpha} \int \frac{dv}{\frac{a_g}{\alpha}+v^2}$$

Then, use the substitution to integrate.

I would get the same, wouldn't I ? One thing I'd like you to tell me. How do you know what subtitution you have to use?

 

[JasonPhysicist]

I would get the same, wouldn't I ? One thing I'd like you to tell me. How do you know what subtitution you have to use?

Not exactly. Notice there's an $$\sqrt \alpha$$ missing on the denominator.

Btw, I know the substitution simply by having done similar integrals many times over. Also, you can get the idea if you remember the usual trigonometric identities.

 

[luuurey]

Not exactly. Notice there's an $$\sqrt \alpha$$ missing on the denominator.

Btw, I know the substitution simply by having done similar integrals many times over. Also, you can get the idea if you remember the usual trigonometric identities.

Got it. So you say it's all about practise, like in normal algebra, right?

 

[JasonPhysicist]

Got it. So you say it's all about practise, like in normal algebra, right?

Exactly. After some time, you'll just get used to it.

 

[Millennial]

It seems quite simple at the first glance, as the equation is separable.

The first thing I'd do is to separate the two variables to get $\displaystyle \frac{dv}{a_g+\alpha v^n}=dt$. From there, you integrate both sides.

The "trick" here is that the left hand side doesn't seem integrable. In fact, it isn't very easy and pleasant to integrate for most n's, but it can be done. I can't find a general algorithm though.

I found a general algorithm to integrate these kinds of functions.

You do a partial fraction decomposition using complex numbers. You'll need the nth roots of unity for this. An example is $\dfrac{1}{1+x^4}$, which can be decomposed using the equality $1+x^4=(x-\sqrt{i})(x+\sqrt{i})(x-i\sqrt{i})(x+i\sqrt{i})$ where $\sqrt{i}=\dfrac{\sqrt{2}}{2}(1+i)$ from Euler's identity. This leaves you with 4 fractions with denominators of degree one. In general, this yields an expression with the hypergeometric function involved, but it can be simplified for integer n to an expression involving only the complex logarithm (or real logarithms + inverse tangent.)