## Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Limits, differentiation, related rates, integration, trig integrals, etc.
FrozenBlood
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### Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

I'm stuck right now on this equation: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

The above equation looks nonlinear, and not separable. At the moment, I can't think of any method to employ when it comes to solving this equation for y, of the few methods I've learned so far up to this point. Again, separating variables didn't work for me, and it doesn't look like the integrating factor method will work either. Plus, it doesn't look exact at all.

Any ideas?

Karl
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### Re: Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Substitute u^3 = y. Hence 3u^2 (du/dx) = dy/dx. Equation becomes

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

Divide out 3u^2 and the equation is linear and can be solved for u in the usual way for linear first order equations. Then backsubstitute u to get y.

FrozenBlood
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Joined: Wed Feb 04, 2009 11:31 pm
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### Re: Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Substitute u^3 = y. Hence 3u^2 (du/dx) = dy/dx. Equation becomes

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

Divide out 3u^2 and the equation is linear and can be solved for u in the usual way for linear first order equations. Then backsubstitute u to get y.
Aren't you the same Karl that runs Karl's Calculus Tutor?

Anyway, I'm just making sure that I've done everything correctly as you said:

u^3 = y
3u^2 (du/dx) = dy/dx
3xu^2 (du/dx) = 6u^3 + 12x^4u^2

Now, the next part, dividing out 3u^2 as you said I should leaves me with:

x(du/dx) = 2u + 4x^2

Dividing both sides by x gives:

du/dx = (2/x)u + 4x
(du/dx) - (2/x)u = 4x

I suppose the next step is to solve for the integrating factor (which I'll let be p(x)):

p(x) = e^integral[(-2/x)] dx
p(x) = e^(-2ln|x| + C)
p(x) = Ce^ln(1/x^2)
p(x) = C/x^2

Multiplying the ODE by this integrating factor (omitting the C since it's sort of extraneous anyway, so now it's just x^-2) gives me:

(x^-2)(du/dx) - (2x^-3)u = 4x^-1

The left side can be converted to:

(d/dx)[ux^-2] = 4x^-1

Multiplying both sides by dx:

d(ux^-2) = 4x^-1 dx

Integrating both sides:

ux^-2 = 4ln|x| + C

Dividing both sides by x^-2 leaves:

u = 4x^2ln|x| + Cx^2
u^3 = (4x^2ln|x| + Cx^2)^3
y = (4x^2ln|x| + Cx^2)^3

That should be the solution, from what I calculated.

Right?

Karl
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### Re: Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Yes I am the same Karl as on KCT.

In my previous post on this thread I got as far as

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

where u^3 = y

Dividing out 3u^2:

x (du/dx) = 2u + 4x^4

You seem to have miscopied the last exponent in your solution.

This equation is linear. To put it into standard form, divide by x and rearrange:

du/dx - 2u/x = 4x^3

Your integrating factor appears to be correct. So other than a clerical mistake, you appear to have it nailed. Run your steps with the correction, then try putting the resulting solution back into the original equation to see if checks out.

FrozenBlood
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Joined: Wed Feb 04, 2009 11:31 pm
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### Re: Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Yes I am the same Karl as on KCT.

In my previous post on this thread I got as far as

3xu^2 (du/dx) = 6u^3 + 12x^4u^2

where u^3 = y

Dividing out 3u^2:

x (du/dx) = 2u + 4x^4

You seem to have miscopied the last exponent in your solution.

This equation is linear. To put it into standard form, divide by x and rearrange:

du/dx - 2u/x = 4x^3

Your integrating factor appears to be correct. So other than a clerical mistake, you appear to have it nailed. Run your steps with the correction, then try putting the resulting solution back into the original equation to see if checks out.
I probably should've done the work on paper instead of typing it all out; maybe then, I could have avoided that careless error early on.

But yes, now I see my mistake. I'll continue from (du/dx) - (2/x)u = 4x^3:

x^-2(du/dx) - (2x^-3)(u) = 4x
(d/dx)(ux^-2) = 4x
d(ux^-2) = 4x dx
ux^-2 = integral(4x)dx
ux^-2 = 2x^2 + C
u = 2x^4 + Cx^2
u^3 = (2x^4 + Cx^2)^3
y = (2x^4 + Cx^2)^3

Check:

x(dy/dx) = 6y + (12x^4)[y^(2/3)]
3x[(2x^4 + Cx^2)^2](8x^3 + 2Cx) = 6[(2x^4 + Cx^2)^3] + (12x^4)(2x^4 + Cx^2)^2
Next step: factoring out (2x^4 + Cx^2)^2 from each side:
(2x^4 + Cx^2)^2[3x(8x^3 + 2Cx)] = (2x^4 + Cx^2)^2[6(2x^4 + Cx^2) + 12x^4]
Dividing both sides by (2x^4 + Cx^2)^2 leaves:
3x(8x^3 + 2Cx) = 6(2x^4 + Cx^2) + 12x^4
24x^4 + 6Cx^2 = 12x^4 + 6Cx^2 + 12x^4
Subtracting 6Cx^2 from both sides leaves:
24x^4 = 12x^4 + 12x^4
24x^4 = 24x^4

So everything checks out.

Karl
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### Re: Differential Equations: x(dy/dx) = 6y + (12x^4)[y^(2/3)]

Excellent work.

The more general lesson here is that any time you have an equation in the form of:

$p(x)\frac{dy}{dx} = q(x)y + r(x)y^s$

the way to attack it is to substitute

$u^{\left(\frac{1}{1-s}\right)} = y$

so that

$\left(\frac{1}{1-s}\right)u^{\left(\frac{s}{1-s}\right)}\left(\frac{du}{dx}\right) = \frac{dy}{dx}$

When you make this substitution, you will always be able to divide $u^{\left(\frac{s}{1-s}\right)}$ out of the equation to make it into a first-order linear. The only exception is when $s=1$, but then the equation is already linear.