Question

Separable differential equation problem

Answer 1

Cool shoot.

Answer 2

after integrating both sides i get\[-y^{-1}=3x^{2}+C\] now i need to use the initial value \[y(1)=\frac{1}{25}\] to find C so i took everything to the -1 and got\[-y=\frac{1}{3}x^{-2}+C\] \[y = -\frac{1}{3}x^{-2}+C\] C stays the same as it is a constant and i"m not doing anything using variables to it

Answer 3

the thing is that when I do it this way (above) as in inverting everything first and then substituting the initial value in, i get one answer but if i don't invert and sub in, and then invert i get a different answer. hope that is clear. please help

Answer 4

@Kainui

Answer 5

i have the answers, so i'm not looking for that. looking for an explanation for the inverting problem i'm having

Answer 6

@TuringTest @satellite73 @estudier

Answer 7

You inverted your equation incorrectly. The answer I got for C is -28.

Answer 8

If you invert it you will multiply y by both sides and then divide the equation by 3x^2+C which isn't what you did.

\[-\frac{ 1 }{ 3x^2+C }=y\]

Answer 9

Now you can plug C back into this equation to have a nice little equation of y without any constants in it.

Answer 10

okay so i see what you did there by why couldn't i do what i did which was to take each term to the -1 ? I feel like i'm making a rookie error...if you take one side to a power you have to take the whole of the other side to that power not each term individually...yeah....thanks :)

Answer 11

Haha no problem I make simple mistakes all the time when I'm caught up in all the partial derivatives and integration and what not. =P

Answer 12

thanks :)

Answer 13

A good way to maybe remember that is to consider what would happen if you solved for the hypotenuse of the pythagorean theorem. a+b=c would be weird lol.