Question

solve the separable differential equation:
dy/dx= 2xycos(x^2)

Someone please tell me how!! I know I have to integrate both sides but that's about all I know. solve the separable differential equation:
dy/dx= 2xycos(x^2)

Someone please tell me how!! I know I have to integrate both sides but that's about all I know. @Calculus1

Answer 1

dy/dx= 2xycos(x^2)
=> dy/y = 2x cos(x^2) dx

Now integrate.

Answer 2

wait did you really bring the dx and y's to the other sides? I can now integrate the right side to be sin(x^2) but does integrating dy/y still cancel out effectively?

Answer 3

orrr is it ln(y)?

Answer 4

Yes,
\[ \int \frac{dy}{y} = \ln y + C \]
Make sure you add a constant to one side.

Answer 5

I know this is just algebra stuff now, but that's what I always get confused on. How do you get from ln(y) + C = sin(x^2) to what I see as the answer: \[Ce^\sin(x^2)\]
Is this correct:
\[e ^{\ln(y) + C} = e ^{\sin(x^2)}\]
and then e and ln(y) cross out to just being y, but how do I justify putting the C on the other side?

Answer 6

exp[ln y + C] = y.e^c

So y = A exp( sin x^2),
where A = e^-C,
that's how.

Answer 7

Wait I didn't follow that at all. Could you re word possibly?

Answer 8

\[e^{\ln y + C} = e^{\ln y}.e^C = y.e^C \]
Yes?

Answer 9

Hence if
ln y + C = sin x^2, then

y . e^C = e^(sin x^2)

i.e.,

\[ y = e^{-C} e^{\sin x^2} \]
\[ \ \ \ = A e^{\sin x^2} \]
where A = e^-C

Answer 10

ohhh wow you just replace e^-c with another variable for a constant. OK! well thank you soooooo much for your help! I'm teaching myself calculus this year because of a lack of course offerings at my school...so thank you for being my supplementary teacher!