Thread: Differential Equations, what are they? a few ?'s

1. Differential Equations, what are they? a few ?'s

I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

first one:
show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

second:
1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
3) think of a solution NOT a meber of the family in part 2
4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

All help appreciated

2. Originally Posted by deltemis
I'm having a tough time figuring out exactly what differential equations are. after reading the chapter on it 3 times over, I do not understand how one equation is a solution of the diff.eq. of some other equation. Can somebody explain to this to me? I've got 2 problems that I found that would help me learn what it is if somebody can explain.

first one:
show that y = x - x^-1 is a solution of the differential equation xy' + y = 2x

there's only 1 example in my book and after following its same steps, i get y' = (1/x^2) + 1, how is that a solution of xy' + y = 2x?

second:
1) what can be concluded about a solution of the equation y' = -y^2 just by looking at it?
2) verify all members of family y = 1/(x+C) are solutions of y' =-y^2
3) think of a solution NOT a meber of the family in part 2
4) find a solution of the initial-value problem: y' = -y^2 y(0)=.5

the second one really has me wondering of how to approach y' = -y^2 and especially in 2), how is that function a solution?

All help appreciated
$\displaystyle x\,\frac{dy}{dx} + y = 2x$

Can you see that the LHS is a product rule expansion of $\displaystyle \frac{d}{dx}\,(x\,y)$?

So $\displaystyle \frac{d}{dx}\,(x\,y) = 2x$

$\displaystyle x\,y = \int{2x\,dx}$

$\displaystyle x\,y = x^2 + C$

$\displaystyle y = x + Cx^{-1}$.

If $\displaystyle C = -1$ then $\displaystyle x - x^{-1}$ is a solution.

3. The derivative of $\displaystyle y = x - x^{-1}$ with respect to $\displaystyle x$ is $\displaystyle y' = 1 + \frac{1}{x^2}$.

Substitute $\displaystyle y = x - x^{-1}$ and $\displaystyle y' = 1 + \frac{1}{x^2}$ in $\displaystyle xy' + y = 2x$. Simplify to an obviously true statement, such as $\displaystyle 0 = 0$.

4. wow thank you both for fast and simplified answers, I hadn't noticed that product expansion and I wouldn't have even thought of substitution

I get it now, thank you both very much