# Thread: Can't figure this one out

1. ## Can't figure this one out

I have no clue where to start with this any advice would be really appreciated. I tried using u-substitution but that did not work.

Solve the initial value problem

dy/dx = 1-sin(5x)/cos(5x)+5x+6 (x>-1), y=3 when x=0

2. ## Re: Can't figure this one out

You need to use paretheses. Is this what you mean?

$\dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6$

or did you mean:

$\dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}$

Assuming it is the latter:

Let $u = \cos(5x)+5x+6$. Then $du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx$

This gives:

$\displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C$

Plugging in the initial values:

$3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)$

This gives $y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3$

4. ## Re: Can't figure this one out

Originally Posted by alexpasty2013
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5. ## Re: Can't figure this one out

Here it is reformatted. Let me know if this is better.

Originally Posted by SlipEternal
You need to use paretheses. Is this what you mean?

$\displaystyle \dfrac{dy}{dx} = 1-\dfrac{\sin(5x)}{\cos(5x)}+5x+6$

or did you mean:

$\displaystyle \dfrac{dy}{dx} = \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}$

Assuming it is the latter:

Let $\displaystyle u = \cos(5x)+5x+6$. Then $\displaystyle du = \left(5-5\sin(5x)\right)dx = 5\left(1-\sin(5x)\right)dx$

This gives:

$\displaystyle \displaystyle y = \int \dfrac{1-\sin(5x)}{\cos(5x)+5x+6}dx = \dfrac{1}{5}\int u^{-1}du = \dfrac{1}{5}\ln\left|\cos(5x)+5x+6\right| + C$

Plugging in the initial values:

$\displaystyle 3 = \dfrac{1}{5}\ln(7)+C$ shows $C = 3-\dfrac{1}{5}\ln(7)$

This gives $\displaystyle y = \dfrac{1}{5}\ln \left| \dfrac{\cos(5x)+5x+6}{7} \right|+3$