Question

SIMPLE DIFFERENTIAL EQUATION: Solve the differential equation dy/dx= y^2/x^3 for y = f(x) with the condition y(1) = 1.

Answer 1

do you know how to "separate the variables" i.e. move all the x's (and dx's) on one side and the y;s on the other ?

Answer 2

yes. I did it, does this look ok?

Answer 3

dy/dx=y2(1/x3)
1/y2 dy= 1/x3 dx
@phi

Answer 4

ok, and if you write it with negative exponents
\[ y^{-2} \ dy = x^{-3} \ dx \]
now integrate both sides
\[ \int y^{-2} \ dy = \int x^{-3} \ dx \]

Answer 5

-y^-1 = -1/2x^-2 + C ? @phi

Answer 6

yes. you can probably clean it up a bit

Answer 7

remember y(1)= 1
that means x=1, y=1 is a solution. So you can solve for C

Answer 8

I'm having a hard time getting the y truly by itself. Can you show me that?

Answer 9

I would not bother. we know the point (1,1) is a solution
(that is what f(1)= 1 means)
so replace x with 1 and y with 1
solve for C

Answer 10

-1^-1 = -1/2(1)^-2 + C
-1= -0.5 + C
-0.5=C

Answer 11

ok.
\[ - \frac{1}{y}= - \frac{1}{2x}- \frac{1}{2} \]
I would multiply both sides by -1 to get rid of all the minus signs
\[ \frac{1}{y}= \frac{1}{2x}+ \frac{1}{2} \]
Next, add the two fractions on the right side

Answer 12

that should be x^2 (not x) in the bottom

Answer 13

the last step would be to "flip" i.e. invert both sides to get
y = some fraction of x

Answer 14

so what does C equal?

Answer 15

C is the unknown constant of integration. You found it equal to -0.5 i.e. - ½
we put in -½ in to your solution
now we simplify /rewrite to make it look a bit nicer.

Answer 16

Ok great! Thanks so much. I have some more questions I would like you to check. I'll post them separately. I'll fan you.

Answer 17

in other words, your solution is
\[ - \frac{1}{y}= - \frac{1}{2x^2}- \frac{1}{2} \\
\frac{1}{y}= \frac{1}{2x^2} +\frac{x^2}{2x^2}\\
y = \frac{2x^2}{x^2+1}
\]

Answer 18

Yep!

Answer 19

Thanks again.