1. ## inital value-problem

how would you solve this initial-value problem?

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{sin(4x)}{3+cos(4x)}$ when $\displaystyle y=5$ and $\displaystyle x=0$

2. ## Re: inital value-problem

Integrate both sides with respect to x. Then use the initial value pair to decide the value of the constant of integration.

3. ## Re: inital value-problem

Originally Posted by Orlando
how would you solve this initial-value problem?

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{sin(4x)}{3+cos(4x)}$ when $\displaystyle y=5$ and $\displaystyle x=0$
$\displaystyle \int dy=\int \frac{\sin(4x)}{3+\cos(4x)} \,dx$

to solve this integral use substitution :

$\displaystyle \cos (4x) = u \Rightarrow du = - 4\sin(4x) dx$

4. ## Re: inital value-problem

so

$\displaystyle y=\int\frac{-4sin(x4)}{3+cos(4x)}dx$

$\displaystyle y=-\frac{1}{4}\int\frac{-4sin(4x)}{3+cos(4x)}dx$

so
$\displaystyle y=-\frac{1}{4}ln(3+cos(4x))+c$

so

$\displaystyle 5=-\frac{1}{4}ln(3)+c$

so

$\displaystyle c=5+\frac{1}{4}ln(3)$

Im I on the right track?

5. ## Re: inital value-problem

Since $\displaystyle ~\cos(0)=1~$ fourth line should be :

$\displaystyle 5= \frac{-1}{4} \cdot \ln(4) + C$

6. ## Re: inital value-problem

sorry silly mistake

so $\displaystyle y=5-\frac{1}{4}ln(\frac{3+cos(4x)}{4})$