# General solution

#### morbello

im asked to find the general solution to the equation below and after find the implicit form ive done some work on it and just wanted to see if im going in the right direction.

the equation

dy$$/dx = e^-3y cos x (1+sin x )^2$$

My attempt at a general solution

1$$/3 e^3 = (sin x +1)^3 = y =e^3/(sin + 1)^3 +c$$

and the implicit form as

1$$/12 (1+sin x )^3/exp (3y)$$

this is harder maths than ive done so not as sure as with most.

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#### HallsofIvy

im asked to find the general solution to the equation below and after find the implicit form ive done some work on it and just wanted to see if im going in the right direction.

the equation

dy$$/dx = e^-3y cos x (1+sin x )^2$$
Do you mean
$$\frac{dy}{dx}= e^{-3y}cos x(1+ sin x)^2$$?

My attempt at a general solution

1$$/3 e^3 = (sin x +1)^3 = y =e^3/(sin + 1)^3 +c$$
Surely you don't mean this- it makes no sense. That says that y is a constant, that it is equal to (sin x+ 1)3, and that it is equal to e3 over a constant!

and the implicit form as

1$$/12 (1+sin x )^3/exp (3y)$$

this is harder maths than ive done so not as sure as with most.
That last is not even an equation!
If you meant
$$\frac{dy}{dx}= e^{-3y}cos x(1+ sin x)^2$$
then it separates into
$$e^{3y}dy= cos x (1+ sin x)^2 dx[/itex] The left side integrates easily and the substitution u= 1+ sin x makes the right side simple. #### morbello no what i ment that y was equaled to e^3/(sinx+1)^3 +c Last edited: #### Shooting Star Homework Helper im asked to find the general solution to the equation below and after find the implicit form ive done some work on it and just wanted to see if im going in the right direction. the equation dy[tex]/dx = e^-3y cos x (1+sin x )^2$$

My attempt at a general solution

1$$/3 e^3 = (sin x +1)^3 = y =e^3/(sin + 1)^3 +c$$
The second eqn doesn't follow from the first. How did you get it? Show the calculation.

Look again at where you got the problem from. What HallsofIvy says makes the most sense.

"General solution"

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