Ordinary Differential Equations problems

• Feb 15th 2009, 10:40 AM
JonathanEyoon
Ordinary Differential Equations problems
I'm having trouble figuring out which method I should be using on some of these problems. Any help in figuring out which methods would be best is much appreciated.

1. (x^3 + y^3)dx - (3xy^2)dy = 0

2. xdy - ydx = sqrt(x^2 - y^2)dx

Just these two for now. Thanks in advance.

Edit: I tried the first one with Homogeneous method but I get to a point after substitution I can't integrate.
• Feb 15th 2009, 11:52 AM
ben.mahoney@tesco.net
the first one is an exact ode
• Feb 15th 2009, 11:58 AM
Chris L T521
Quote:

Originally Posted by JonathanEyoon
I'm having trouble figuring out which method I should be using on some of these problems. Any help in figuring out which methods would be best is much appreciated.

1. (x^3 + y^3)dx - (3xy^2)dy = 0

2. xdy - ydx = sqrt(x^2 - y^2)dx

Just these two for now. Thanks in advance.

Edit: I tried the first one with Homogeneous method but I get to a point after substitution I can't integrate.

The first one is equivalent to $\frac{\,dy}{\,dx}-\frac{1}{3x}y=\tfrac{1}{3}x^2y^{-2}$, which is a Bernouli equation. Make the substitution $z=y^{1-n}=y^{1-(-2)}=y^3\implies y=z^{\frac{1}{3}}$. Thus, $\frac{\,dy}{\,dx}=\frac{\,dy}{\,dz}\frac{\,dz}{\,d x}=\tfrac{1}{3}z^{-\frac{2}{3}}\frac{\,dz}{\,dx}$

The DE then becomes $\tfrac{1}{3}z^{-\frac{2}{3}}\frac{\,dz}{\,dx}-\frac{1}{3x}z^{\frac{1}{3}}=\tfrac{1}{3}x^2z^{-\frac{2}{3}}\implies\frac{\,dz}{\,dx}-\frac{1}{x}z=x^2$, which is a linear equaton. Can you take it from here?

For the second one, it is equivalent to $\frac{\,dy}{\,dx}-\frac{y}{x}=\sqrt{1-\left(\frac{y}{x}\right)^2}$

Now apply the substitution $u=\frac{y}{x}\implies y=ux$. Thus, $\frac{\,dy}{\,dx}=u+x\frac{\,du}{\,dx}$

Thus, the DE becomes $u+x\frac{\,du}{\,dx}-u=\sqrt{1-u^2}\implies x\frac{\,du}{\,dx}=\sqrt{1-u^2}$. This is a separable equation. Can you take it from here?

Does this make sense?
• Feb 15th 2009, 12:03 PM
Chris L T521
Quote:

Originally Posted by ben.mahoney@tesco.net
the first one is an exact ode

Its not exact, because $\frac{\partial}{\partial y}\left[x^3+y^3\right]=3y^2{\color{red}\neq}-3y^2=\frac{\partial}{\partial x}\left[-3xy^2\right]$
• Feb 15th 2009, 12:10 PM
JonathanEyoon
Thanks alot. Makes perfect sense.
• Feb 15th 2009, 12:27 PM
JonathanEyoon
Ok got some more. You don't need to solve it out for me since i'm quite familiar with the procedure's of the different methods but it's just rearranging the problems to make it look like any one of the 5 that i'm having trouble with (Speechless). Saw the mistake~! Thanks for fixing it up.

1.(1+x^3)dy - ((x^2)y)dx = 0

2.(1+x)dy/dx - xy = x^3 + x^2

For the first one, That should be separable correct? If so, how about the 2nd?
• Feb 15th 2009, 12:30 PM
HallsofIvy
Quote:

Originally Posted by JonathanEyoon
I'm having trouble figuring out which method I should be using on some of these problems. Any help in figuring out which methods would be best is much appreciated.

1. (x^3 + y^3)dx - (3xy^2)dy = 0

2. xdy - ydx = sqrt(x^2 - y^2)dx

Just these two for now. Thanks in advance.

Edit: I tried the first one with Homogeneous method but I get to a point after substitution I can't integrate.

Yes, these are both "homogeneous". In the first, if we replace x by ax, y by ay, it becomes (a^3x^3+ a^3y^3)dx- 3a^2xy^2dy and canceling the a^3 terms we get just what we started with. That means it can be rewritten as a differential equation for y/x.

Let u= y/x so y= xu. y'= xu'+ u so dy= xdu+ udx. The equation becomes $(x^3+ x^3u^3)dx- 3x(x^2u^3)(xdu+ udx)= 0$
$(x^3+x^3u^3-3x^3u^3)dx- 3x^4u^3du= 0$
dividing through by $x^3$
$(1+ u^3- 3u^3)dx- 3xu du= 0$
which is separable. it separates as
$\frac{1}{x}dx= \frac{u}{1- 2u^3} du$

Is that what you got? The righthand side is tedious but doable. The denominator can be factored by solving $1- 2u^3= 0$. $u^3= \frac{1}{2}$ is a solution so $u= ^3\sqrt{1/2}$.
Now write out $(u-^3\sqrt{1/2})(u^2+ au+ b)= u^3+(a-^3\sqrt{1/2})u^2+ (b- a ^3\sqrt{1/2})u- b;^3\sqrt{1/2}= u^3- 1/2$ to see that we must have $a= ^3;\sqrt{1/2}$ and $b= ^3;\sqrt{1/2}$ so that the denominator factors as $(-1/2)(u^3- 1/2)= (-1/2)(u- ^3\sqrt{1/2})(u^2+ ^3\sqrt{1/2}u+ ^3\sqrt{1/4})$
That quadratic term doesn't factor with real numbers because its zeros are imaginary but we can complete the square: $(u- ^3\sqrt{1/2})^2+ \frac{3;^3\sqrt{1/2}}{4}$.

Now the left hand side can be rewritten as
$\frac{u}{(u- ^3\sqrt{1/2})(u- ^3\sqrt{1/2})^2+ \frac{3;^3\sqrt{1/2}}{4}$
and that can be integrated using partial fractions.

For the second problem if we multiply both x and y by a we get $axdy- aydx= \sqrt{a^2x^2+ a^2y^2}=- a\sqrt{x^2+ y^2}$ and again we can cancel the "a"s so this is a homogeneous equation. Let u= y/x so that y= xu and dy= xdu+ udx and the equation becomes $(x(xdu- udx)- xudx= \sqrt{x^2- u^2x^2}dx$ or $x du= \sqrt{1- x^2}dx$. That is again homogeneous and certainly simpler than the previous problem! It separates as $\frac{du}{\sqrt{1-u^2}}= \frac{dx}{x}$ and both sides are easy to integrate.
• Feb 15th 2009, 12:56 PM
Chris L T521
Quote:

Originally Posted by JonathanEyoon
Ok got some more. You don't need to solve it out for me since i'm quite familiar with the procedure's of the different methods but it's just rearranging the problems to make it look like any one of the 5 that i'm having trouble with (Speechless). Saw the mistake~! Thanks for fixing it up.

1.(1+x^3)dy - ((x^2)y)dx = 0

2.(1+x)dy/dx - xy = x^3 + x^2

For the first one, That should be separable correct? If so, how about the 2nd?

Yes the first one is separable.

The second one is linear, because $(1+x)\frac{\,dy}{\,dx}-xy=x^2(1+x)\implies\frac{\,dy}{\,dx}-\frac{x}{1+x}y=x^2$.

Can you continue?
• Feb 15th 2009, 12:56 PM
JonathanEyoon
Quote:

Originally Posted by JonathanEyoon
Ok got some more. You don't need to solve it out for me since i'm quite familiar with the procedure's of the different methods but it's just rearranging the problems to make it look like any one of the 5 that i'm having trouble with (Speechless). Saw the mistake~! Thanks for fixing it up.

1.(1+x^3)dy - ((x^2)y)dx = 0

2.(1+x)dy/dx - xy = x^3 + x^2

For the first one, That should be separable correct? If so, how about the 2nd?

(Worried) Test on Tuesday and I'm worried
• Feb 15th 2009, 04:39 PM
Chris L T521
Quote:

Originally Posted by JonathanEyoon
(Worried) Test on Tuesday and I'm worried

I already responded to that post... (Speechless)