Solving differential equations.

[STEMucator]

Homework Statement

I came across two, not so obvious DEs that have stumped me abit.

(1) x²y' = xy + x²e^y/x
(2) x²y' + 2xy = 5y³

Homework Equations

I know these are not separable, and more than likely require an integrating factor to put them into exact form so I can integrate them that way.

The Attempt at a Solution

I'm sort of stuck on both of them.

For (1) I divided through by x² to get y' = y/x + e^y/x which further yields y' - y/x = e^y/x which is sadly implicit in nature and not solvable by means of a regular integrating factor.

Its the same for (2), once again I divide through by x² to attain y' + (2/x)y = 5y³/x² which is once again implicit and not solvable by regular means.

I know I probably need to put these into exact form somehow, but I'm having trouble putting them into the form :

$M(x, y) + N(x, y) \frac{dy}{dx} = 0$ So I can solve for an integrating factor $\mu (x)$ which satisfies :

$$\frac{\frac{∂M(x, y)}{∂y} - \frac{∂N(x, y)}{∂x}}{N(x, y)} = \frac{d\mu}{dx}$$

 

[rock.freak667]

For the first one, try a substitution of y=Vx.

 

[voko]

For the second one, observe x^2y' + 2xy = (x^2y)'.

 

[STEMucator]

For the second one, observe x^2y' + 2xy = (x^2y)'.

I didn't know what substitution was, but I learned it quickly and I managed to get the solution to the first equation to be :

-e^-y/x = ln|x| + c which I decided to obviously leave implicit as no real solutions of y would exist.

As for the second equation, voko you mentioned I could re-write this as :

d/dx [x²y] = 5y³ I'm not sure how this would help though?

 

[voko]

What is in square brackets is another function u(x). Can you express y via x and u?

Speaking of the first one, consider the case when c is a large negative constant.

 

[STEMucator]

What is in square brackets is another function u(x). Can you express y via x and u?

Speaking of the first one, consider the case when c is a large negative constant.

So for the first one I COULD re-write it as :

y = xln|d-ln|x|| where d is some arbitrary constant ( I get d after some arithmetic ).

As for the second one, are you hinting that a substitution v(x) = x²y which simplifies to x^-2v(x) = y is what I need here?

 

[voko]

For the first one, I think it should be y = -x ln...

The second one, try :)

 

[STEMucator]

For the first one, I think it should be y = -x ln...

The second one, try :)

For the first one when I integrate and sub back m obvious choice of v = y/x I get :

-e^^-y/x = ln|x| + c
e^^-y/x = d - ln|x|
-y/x = ln|d - ln|x||
y = -xln|d-ln|x||

I c wut u did thur.

As for the second one I get the answer as : y = ±$\sqrt{\frac{x}{2} + c}$

I think that's good?

 

[voko]

I do not think the second result is correct. Please show your steps.

 

[STEMucator]

I do not think the second result is correct. Please show your steps.

x²y' + 2xy = 5y³
d/dx [x²y] = 5y³

Substituting v = x²y yields :

dv/dx = 5y³

Also notice that y = vx^-2 so :

dv/dx = 5(vx^-2)³
dv/dx = 5v³x^-6

Separating variables and integrating gives us :

-(1/2v²) = -x^-5 + c

Subbing back v = x²y and simplifying :

-(1/2(x²y)² = -x^-5 + c
2x⁴y² = x⁵+c
y² = x/2 + c/x⁴
y = ±$\sqrt{\frac{x}{2} + \frac{c}{x^4}}$

I really should write my steps out haha.

 

[voko]

-(1/2(x²y)² = -x^-5 + c

Good so far.

2x⁴y² = x⁵+c

Wrong, both the LHS and the RHS.

 

[STEMucator]

Good so far.

Wrong, both the LHS and the RHS.

Oh snap, silly mistake. I should also stop rushing when showing my work.

-(1/2(x²y)²) = -x^-5 + c
1/(2x⁴y²) = x^-5 - c
2x⁴y² = 1/(x^-5-c)
y² = 1/(2x⁴(x^-5-c))

y = ±$\sqrt{\frac{1}{2x^4(x^{-5}-c)}}$

 

[voko]

That's better :) To be completely sure, you could try and plug that into the original equation.

 

[STEMucator]

That's better :) To be completely sure, you could try and plug that into the original equation.

Yup it works perfectly. Although I DO have a technicality question out of my own curiosity.

Say I was given an initial condition y(x₀) = y₀.

If y₀ > 0, then the solution corresponds to the positive portion of y.

If y₀ < 0, then the solution corresponds to the negative portion of y.

Now if y₀ = 0, I'm not really sure what happens here? Which solution would I pick or would it never be the case that y₀ = 0 because y is discontinuous there regardless of my choice. Seems logical, but some confirmation would be nice.

Thanks for all your help btw :)

 

[voko]

y = 0 is a singular solution in this equation. It corresponds to an infinite c. Note the regular solutions approach it as x goes to zero.

 

[STEMucator]

y = 0 is a singular solution in this equation. It corresponds to an infinite c. Note the regular solutions approach it as x goes to zero.

Ahh I see. That makes sense now, thank you.