# Thread: Solving first order differential equation

1. ## Solving first order differential equation

xy^1/2 dy/dx - xy = -y^2/3 initial condition y(1) = 0

The first thing i did was divide it through by x:

so i got y^1/2 dy/dx - y = (-y^3/2)/x

Then i divided it through by y^1/2

from this I got:

dy/dx - y^1/2 = -y/x

But now im stuck as to how to solve it, Do i use Bernoulli's equation? or am i totally off...?

2. Well, between one or more steps, you've switched the exponent of y on the RHS. Is it 2/3 or 3/2? That will make a HUGE difference.

3. Oh woops! Its suppose to be 3/2 -.-"" must have typed it wrong

4. Ok. Starting from the original DE:

$x\,y^{1/2}\,y'-xy=-y^{3/2}.$

I'm not so sure I would solve for $y'$. You might be able to get a product rule going here. Rearrange:

$x\,y^{1/2}\,y'+y^{3/2}=xy.$

I like the idea of dividing through by $y$, which yields:

$x\,y^{-1/2}\,y'+y^{1/2}=x.$

Now the LHS is close to a product rule, but not quite. What's missing?

5. uumm What do you mean by Product rule..?

6. I just mean the normal product rule from differential calculus:

$(fg)'=f'g+fg'.$

7. I didn't know we could solve differential equations using the product rule...also I'm not quite seeing how the LHS looks like a product rule apart from having xy^-1/2

8. Sure you can. You can also use the quotient rule. The idea is this: if you can write the LHS as a product rule (fg)', then you can integrate both sides directly, because the LHS is then a total derivative. So, for example, suppose you had to solve the DE

$2xyy'+y^{2}=e^{x}.$

You can "notice" that the LHS is a total derivative by seeing that

$(xy^{2})'=2xyy'+y^{2}.$ That means you can rewrite the DE as follows:

$(xy^{2})'=e^{x}.$ Integrating both sides yields

$\displaystyle{\int (xy^{2})'\,dx=\int e^{x}\,dx,}$ and so

$xy^{2}=e^{x}+C,$ by the Fundamental Theorem of the Calculus. Solve for $y$ and you're done.

In terms of your DE, you've almost got a product rule, but not quite. If you look at

$(x\sqrt{y})'=\frac{1}{2}\,x\,y^{-1/2}\,y'+y^{1/2},$

you'll see that the entire LHS is very close to being the derivative of a product. The problem is that the coefficients of each term are not the same in this product, whereas the coefficients are the same in your DE. Any ideas on how to fix this? This is a fantastic application of the problem-solving strategy of introducing symmetry where there is none.

9. Thanks for you help I have solved the question already

10. Great! Have a good one.