Question

if dy/dx= (x^2)(y^2 ), then d^2y/dx^2 = ....

Not sure if I should use differiantial equation to solve or product rule.
Help please!!!

Answer 1

\[\large y'=x^2 y^2\]
Find \(\large y''\)

Product rule \c:/

Answer 2

Do you know the final answer?

Answer 3

Do I know the answer? :O yes.. but try to work it out for yourself silly!

Here is the setup for product rule.

\[\large y''=\color{royalblue}{(x^2)'}y^2+x^2\color{royalblue}{(y^2)'}\]

Need to take the derivative of the blue terms.

Answer 4

I got 2xy^2/ 1-(2y)(x^2) = dy/dx

Answer 5

I'm not sure why you have a divide by 1 in the middle, that's kinda confusing.
Is this what you came up with?
We'll need to make a minor fix if so.
\[\large y''=2xy^2-2yx^2\]

Answer 6

yes that is what I got when I did product rule.

Answer 7

But don't we have write dy/dx for the derivative of 2y?

Answer 8

Yes very good, we have to multiply the 2y by dy/dx.

Since we took our derivative `with respect to x`,
Whenever we take the derivative of y, we have to connect a dy/dx term to it.

Answer 9

\[\large y''=2xy^2+2yx^2y'\]

Another small error, it should be addition in the middle. I'm not sure why it changed to subtraction when you took the derivative. hmm

Answer 10

Depending on what your teacher wants ~ You may be asked to simplify this down a little bit further.

Remember that our y' was given to us to be,\[\large \color{orangered}{y'=x^2 y^2}\]

We can substitute this in for our y' that showed up if we want to simplify it down further.
\[\large y''=2xy^2+2yx^2\color{orangered}{y'}\]

Answer 11

got it! 2x^4y^3+ 2xy^2

Answer 12

yay good job! c:

Answer 13

Thanks.