Solve for these 2 ODE equations How do I solve these ODE's? Do I need to find the value of my y variable or do I need to find solutions for the function in the derivative? Would like to know how it is done. Many thanks :). P.S. new to all these higher level mathematics.
The first ODE

Let's write it in a more convenient notation:
$$yy'=0,\,\, y(0)=1$$

What does this ODE says? a function times it's derivative is the constant zero, so at least one of the functions should be the the zero function. One trivial solution is $y=0$ since it satisfies the ODE, but it doesn't satisfy the initial value! So we must have $y'=0$ so $y=c$, from the initial value we conclude the solution is $y=1$.

The second ODE

You can solve this ODE using the Laplace transform. let me know in the comments if you don't know how.



SKETCH FOR SOLUTION USING THE LAPLACE TRANSFORM

We usually solve ODEs by applying the Laplace transform for linear differential equations (not necessarily homogenous!), most popularly of order 2 or less. 

The most useful fact about  the Laplace transform, the one that makes it useful to solve linear ODEs is the following relation between the Laplace transform and the derivative:
$$\mathcal{L}(f')=s\mathcal{L}(f)-f(0)$$

Do note we also need and an initial value of $f$ evaluated at $0$.
Another useful fact about the Laplace transform- since it's defined by the integral:

$$\mathcal{L}(f)=\int_{0}^\infty f(t)e^{-st}dt$$

It is a linear transformation! (This can be seen by using the linearity of the integral). 

These properties of the Laplace transform actually reduce solving an IVP from a differential problem to an algebraic one (It is is a really rich field of research called Operational Calculus!). After applying the transform to both sides of the equation we re-arrange the equation to that we can use the last important property of the Laplace transformation: It is invertible! so after re-arranging we can apply $\mathcal{L}^{-1}$ to both sides and thus solve for $f$ .

Table of the laplace transform for common functions
For your first equation we get
$$y(x)=0$$ or $$y'(x)=0$$
The second one has the solutions
$$y(x)=c_2 \sin (2 x)+c_1 \cos (2 x)+\dfrac{1}{3}\sin x$$