Solving Differential Equations

[mrmonkah]

Homework Statement

Solve the following: xdx = y^{2}dy

Homework Equations

Fundamental theorem of calculus - thanks gabbagabbahey

The Attempt at a Solution

\frac{dy}{dx}=\frac{x}{y^{2}}

\int\frac{dy}{dx} = \int\frac{x}{y^{2}}

=\frac{x^{2}}{2y^{2}}

So with the question, I've integrated both sides to find y^2 on the bottom, so should i rearrange to find y on the LHS and the x on the RHS?

 

[gabbagabbahey]

Homework Equations

NA

Unless NA is your abbreviation for the fundamental theorem of calculus, you are mistaken

The Attempt at a Solution

\stackrel{dy}{dx} = \stackrel{y^{2}}{x}

What happens if you integrate both sides of the equation x=y^2\frac{dy}{dx} with respect to x?

 

[mrmonkah]

Hi gabbagabbahey,

Sorry, i am just getting to grips with the funky math features on the site, so my translation from paper to web isn't vry good. First of all i put in a monster mistake on the web, working on correcting this now.

 

[mrmonkah]

Okay, more or less fixed now... thanks again Gabb

 

[mrmonkah]

Right, so carrying on from my first post,

if i rearrange for y, i get:

y = \sqrt{\frac{x^{2}}{2}}

 

[Char. Limit]

Your starting point is

x dx = y^2 dy

Right?

You can just integrate both sides right there. Start with that.

 

[mrmonkah]

Ok, so if i integrate both sides i get:

\frac{x^{2}}{2} = \frac{y^{3}}{3}

and y = \sqrt[3]{\frac{3x^{2}}{2}}

Surely this isn't right is it? I don't recall coming across cubic roots in 'this particular' module. (I am simply looking for familiarity with the rest of the course)

 

[Char. Limit]

You forgot the plus C.

 

[mrmonkah]

Ahh okay fair enough Char.Limit, i am confused as to why re-arranging the initial equation (as i did earlier) yields such a different result?

 

[Char. Limit]

It's mainly because you can't integrate x/y^2 dx, because y also depends on x.

So you can see why you couldn't integrate...

\int \frac{x}{y^2(x)} dx

 

[mrmonkah]

Oh i see now. So with questions like these, i should generally keep all the y's on one side and the x's on the other? Ill attempt another question and post it to see if i have my head in the right place. Thank you Char.Limit, and as ever, you make a good point.

 

[Char. Limit]

No problem. Glad I could help.

 

[mrmonkah]

Homework Statement

Solve the following: \frac{dy}{dx} = \frac{1 + y}{1 + x}

Homework Equations

Fundamental theorem of calculus - thanks gabbagabbahey

The Attempt at a Solution

Re-arranging to get y terms and x terms on opposite sides:

\int\frac{dy}{1 + y} = \int\frac{dx}{1 + x}

Which gives:

ln(x + 1) = ln(y + 1)

and so:

y = x

Does this work out?

 

[Char. Limit]

Close, but you still need to remember the +C.

 

[mrmonkah]

Damn, every time. A habit i need to get used to.

 

[Char. Limit]

No real problem. You just have to make sure you do it.