Question

dy/dx=6sin^2x cosx, y(0)=6. Solve the initial value problem.

Answer 1

Looks like you could separate variables, and some u-substitution. Familiar with those ideas?

Answer 2

Yes I am familiar with substituting u to find the answer.

Answer 3

Great! So lets start by seperating variables:
\[\frac{dy}{dx} = \sin^2(x)\cos(x)\]
\[1dy = \sin^2(x)\cos(x)dx\]
\[\int 1dy = \int \sin^2(x)\cos(x)dx\]

Answer 4

let \(u = \sin(x)\)
the \(du = \cos(x) dx\)

Answer 5

Alright, so I see that you started out with the chain rule and then you substituted u

Answer 6

Yah.... chain rule :)

Answer 7

Alright, I'm following you. You got 3sin^2x from dividing 6/2. So does sin^2x turn into (1-cos^2x)?

Answer 8

I wasn't leaning toward using any identity, but you could.... Note:
\[\int 1\; dy = \int \sin^2(x) \cos(x) dx = \int u^2du\]
by substitution. So just power rule.

Answer 9

Okay

Answer 10

The left hand side \(\int 1 dy = y\)
The right hand side \(\int u^2 du = \dfrac{1}{3} u^3 + c = \dfrac{1}{3}\sin^3(x) + c\)

Answer 11

\[y = \frac{1}{3}\sin^3(x) + c\]
The rest of the problem is figuring out what c is.