ODE, bernoulli equation -> leads to crazy integral

[Jonnyb42]

ODE, bernoulli equation --> leads to crazy integral !

Homework Statement

An Initial Value Problem, ODE (Bernoulli equation)

ODE: [x^2]*y' + 2*[x^3]*y = [y^2]*(1+2*[x^2])
IV: y(1) = 1/2

Homework Equations

general form of Bernoulli's equation:
y' + a(x)y = b(x)*[y^n]

First order, linear ODE form:
y' + a(x)y = b(x)

The Attempt at a Solution

I begin by simplifying the Bernoulli-type equation into a linear form, at which point I end up with an integral that I DO NOT KNOW HOW TO SOLVE.

I have either really messed up somewhere along the way to that Integral, or not and I just dont' know how to solve it.

Thanks for any help.

 

[tiny-tim]

Hi Jonnyb42!

Looks ok to me …

but you can't solve that integral unless you're allowed to use the "error function" erf(x).

 

[Tomer]

No further ideas here, sorry :-\

 

[HallsofIvy]

Are you required to actually integrate it? I remember in my first D.E. class, I had one problem in the text that I could reduce to an integral but could not integrate. If finally gave up and looked in the back of the book where I found that the answer was given in terms of that integral!

 

[Jonnyb42]

But how could I get it even in terms of the integral? I need a constant of integration to solve for so y(1) = 1/2

 

[ehild]

Substitute z=x/y. You will get an easy separable differential equation.

ehild

 

[ehild]

With the substitution z=x/y,

y=x/z, y'=(z-xz')/z^2,the ODE becomes

x^2(z-xz')/z^2+2x^4/z=(1+2x^2)x^2/z^2.

Multiplying with z^2/x^2, and rearranging the equation:

z-xz'+2x^2 z=(1+2x^2), which is a separable ODE:

xz'=(1+2x^2)(z-1)--->z'/(z-1)=(1+2x^2)/x

ehild

 

[Jonnyb42]

ehild, thank you!

How do you know to do that substitution?

thanks again,
Jonny

 

[ehild]

I did not know that substitution, I just tried...

Dividing both sides of the equation by y^2*x^2, it becomes

y'/y^2+ 2*(x/y) = 1/x^2+2.

So why not try to choose x/y the new variable? The derivative of x/y will contain the term y'/y^2...

There are no recipes for every ODE-s, and there can be different methods for the same ODE. The recipe suggested the substitution z=1/y, but it leads to a difficult integral. You have to be creative

ehild

 

[Jonnyb42]

thanks very much!

jonny

 

[tiny-tim]

oops!

Hi Jonnyb42!

Looks ok to me …

but you can't solve that integral unless you're allowed to use the "error function" erf(x).

oops!

your original method was fine , and on careful inspection your final integral

∫ -e^-x2(2 + 1/x²) dx​

is simply e^-x2/x + C

 

[ehild]

your original method was fine , and on careful inspection your final integral

∫ -e^-x2(2 + 1/x²) dx​

is simply e^-x2/x + C

Ingenious! How can one find out this integral?

ehild

 

[tiny-tim]

just by looking hard enough …

if it's (fe^-x2)',

then that's (f' -2xf)e^-x2, so -2xf = -2 works. ie f = 1/x, f' = -1/x²

 

[ehild]

You can look really hard.
Those who do not have such eyes would input it to Wolframalpha...

ehild

 

[Dickfore]

It's an initial value problem. So, if you have a linear ODE:
<br /> z&#039; + a(x) z = b(x)<br />
and you find the integrating factor:
<br /> \mu(x) = e^{\int{a(x) \, dx}}<br />
without caring about the arbitrary integrating constant, because that would give an arbitrary multiplicative factor to your integrating factor, then, the solution to the initial value problem with:
<br /> z(x_{0}) = z_{0}<br />
is:
<br /> \mu(x) \, z(x) - \mu(x_{0}) \, z_{0} = \int_{x_{0}}^{x}{\mu(t) \, b(t) \, dt}<br />

From here you can immediately see that if \mu(x) gets multiplied by an arbitrary constant, each term in the above expression gets multiplied by it and the equality holds.