2 differentials, one of which is almost solved

[bennyska]

Homework Statement

so, problem 1:
xy' + 2y = 6x²y^1/2

Homework Equations

so this is a bernoulli, where, in the form y' + p(x)y = q(x)yⁿ

The Attempt at a Solution

xy' + 2y = 6x²y^1/2
y' + (2/x)y =6xy^1/2
so in the form, p(x) = 2/x, q(x) = 6x, and v = y^1/2
dividing both sides by y^1/2
y^-1/2y' + (2/x)y^1/2 = 6x

substitute v = y^1/2 and simplify
this becomes:
2dv/dx + (2/x)v = 6x
dv/dx + (1/x)v = 3x
this is linear, and let mu(x)= e^{int. 1/x dx} = e^{ln x} = x
multiply both sides, and get
d/dx(xv) = 3x²
integrate
xv = x³ + c0
v = x² + c0/x
replace v
y^1/2 = x² + co/x
y = (x² + c0/x)²

here's the problem. when i do this without the constant of integration, it works fine, i.e. (x²)² is a valid solution. so I've gone over this several times, but can't find my mistake. any help?

problem no 2: i can't get started. a point in the right direction would be appreciated.

e^x+ y*e^xy + (e^y+ x*e^yx)y'=0

as i type it, i wonder if i copied something down wrong, having the yx in reverse alphabetical order. oh well, yeah, a pointer would help. thanks!

 

[andylu224]

You did nothing wrong for the first problem. Just try plugging the terms into the original ODE again carefully.

For the 2nd problem, make y' the subject and think about what happens when something is implicitly differentiated. I think the answer is e^y + e^x + e^xy = 0.

Good luck.

 

[vela]

Your answer to the first problem is correct. Perhaps your error occurs while trying to check your answer.

In the second problem, you can rewrite it slightly:

(e^x+ye^{xy})dx+(e^y+xe^{xy})dy=0

Does this form jog your memory?

 

[bennyska]

how about one more small hint?

 

[vela]

It's an exact differential. You can see this because

\frac{\partial}{\partial y}(e^x+ye^{xy}) = \frac{\partial}{\partial x}(e^y+xe^{xy})

So there's some function \Phi(x,y) such that

d\Phi = (e^x+ye^{xy})dx+(e^y+xe^{xy})dy

 

[bennyska]

i seem to remember my last class skipping over this section, but i have the book and i'll see what i can do. thanks.