# Thread: 10.09.15t solve the intial value problem

1. ## 10.09.15t solve the intial value problem

$\tiny{10.09.15t}$
$\textrm{ solve the intial value problem}$
\begin{align*}\displaystyle
\theta \frac{dy}{dx}+y&=\cos \theta,>0; \, y(\pi)=1 \\
&=
\end{align*}

ok I'm not in a class right now so trying to self teach but ????

2. ## Re: 10.09.15t solve the intial value problem

Are you sure it's $$\theta \frac{dy}{dx} + y=\cos(\theta)$$ and not $$\theta \frac{dy}{d\theta} + y = \cos (\theta)$$

3. ## Re: 10.09.15t solve the intial value problem

yeah you are correct ..... typo

4. ## Re: 10.09.15t solve the intial value problem

No problem

So for $$\theta\frac{dy}{d \theta} + y = \cos (\theta)$$
Write as
$$\frac{d}{d\theta} \left( \theta y\right) = \cos (\theta)$$
(Do you know where this comes from?...)

Now, integrate both sides with respect to $\theta$. What do you get?

5. ## Re: 10.09.15t solve the intial value problem

$\frac{d}{d\theta} \left( \theta y\right) = \cos (\theta)$

is it

$y=-sin(\theta)$

6. ## Re: 10.09.15t solve the intial value problem

If you were to apply the product rule to $\frac{d}{d\theta}(\theta y)$ you would get $y+ \theta \frac{dy}{d\theta}$, that is how I got that.

Next step is to integrate both sides, so

$$\int \frac{d}{d\theta} \left(\theta y\right) d\theta = \int \cos(\theta ) d\theta$$

7. ## Re: 10.09.15t solve the intial value problem

I'm logging off now, so here's a solution, I've assumed you meant $\theta >0$ in your original question.

$$\theta\frac{dy}{d\theta}+y=\cos(\theta),\ \theta >0,\ y(\pi)=1$$
Write as
$$\frac{d}{d\theta}(\theta y) = \cos(\theta)$$
Integrate both sides with respect to $\theta$
$$\int \frac{d}{d\theta}(\theta y) d\theta = \int \cos (\theta)d\theta$$
Which gives
$$\theta y = \sin(\theta) + C$$
C is a constant. Now divide by $\theta$ to isolate $y$,
$$y=\frac{\sin(\theta)+C}{\theta}$$
Applying the initial condition:
$$y(\pi) = \frac{\sin(\pi) + C}{\pi}=1$$
$$\frac{C}{\pi}=1$$
$$C=\pi$$
So the solution is
$$y(\theta)=\frac{\sin(\theta)+\pi}{\theta}$$

much mahalo