What is the solution to dy/dx = (x^2)/y, with the initial condition y(3) = -2

Question

can someone pls show the steps they used to solve this! I keep getting y = sqrt of ((2x^3)/3 -14), but the right answer is supposed to be y = - sqrt of ((2x^3)/3 -14), and I'm not sure where the negative sign came from. thanks :)

Answer 1

dy/dx = x²/y

Separate variables: ydy = x²dx

Integrate both sides: (1/2)y² = (1/3x³ + C

Since y = -2 when x = 3, we have 2 = 9 + C. So, C = -7

y² = (2/3)x³ - 14

y = √[(2/3)x³ - 14] OR y = -√[(2/3)x³-14], but xince y = -2 when x = 3, only the second equation works.

So, y = -√[(2/3)x³ - 14]

Answer 2

Hello Sydney W.,

The reason the negative sign shows up is from the consequence of the square root and the necessity of the initial condition

When you take the square root of (almost) anything, you have to give consideration that the plus or minus of the argument could be squared to get back to the thing prior to the square-rooting (there is probably more exact terms for what I'm saying above but I'm not a mathematician that knows all of those fancy things)

So, the solution would actually look like this for a second:

y = ± √ [ (2 x³ ) / 3 - 14 ]

But then, you need the initial condition to come in since:

y( 3 ) = - 2

Thus, you are allowed to over-rule the ± in favor of the minus to satisfy the initial condition

Thus,

y = - √ [ (2 x³ ) / 3 - 14 ]

I hope this helps! Message me in the comments if you have any questions, comments, or concerns with the work above!