Solve the differential equation: $\frac{d^2 y}{dx^2}+y \frac{dy}{dx}=0$ Solve the differential equation
$$\frac{d^2 y}{dx^2}+y \frac{dy}{dx}=0$$
$$y(0)=1, \quad y'(0)=-1/2$$
Hint: use $w(y(x))=y'(x)$
How can I solve this using the hint specifically? Would it just be $w'+yw=0$?
 A: Let $w=\frac{dy}{dx}$, then
$$\frac{d^2y}{dx^2}=\frac{dw}{dx}=\frac{dw}{dy}\frac{dy}{dx}=w\frac{dw}{dy}.$$
So the given equation can be written as:
\begin{align*}
w\frac{dw}{dy} + yw & = 0\\
\frac{dw}{dy}  & = -y && (\text{assuming } y=\text{constant is already taken care of for the case $w=0$})\\
w & = \frac{-y^2}{2}+C
\end{align*}
Can you proceed from here?
A: The same as in @Anurag A answer the ordinary differential equations of the form $f(y,y',y'')=0$ are solved through considering $y'$ as another variable like $u$ and you obtain:
$$\begin{cases}y'=u\\ y''=\frac{dy'}{dx}=\frac{du}{dx} \end{cases}$$
$$\frac{du}{dx}=\frac{du}{dy}.\frac{dy}{dx}=u\frac{du}{dy}$$
then by replacing $y'$ with $u$ and $y''$ with $u\frac{du}{dy}$ a first order differential equation with respect to $u$ and $y$ is obtained. Therefore, your ODE:
$$y''+y y'=0$$
$$y'(0)=-1/2\quad y(0)=1$$
becomes:
$$uu'+yu=0$$
deviding both sides by $u$ it turns into
$$\frac{du}{dy}=-y$$
which is a separable equation.
More about differential equations of this type is available in Elementary Differential Equations by Richard Boyce and DiPrima.
EDIT: The problem which was mentioned in the comments turns into 
$$2tw'+w=0$$
which is a separable equation:
$$\frac{w'}{w}=\frac{-1}{2t}$$
integration gives you:
$$w=c_1t^{1/2}$$
Now integration from $w$ (which is $y'$) with respect to $t$ gives you the answer:
$$y = 2c_1t^{1/2}+c_2$$
which satisfies the equation.
The difference between the two methods is that the first one contains no independent variable like $t$ but the second one does.
A: You have 
$$y''(x)+y(x)y'(x)=0$$
and by the hint $w(y(x))=y'(x)$ we also have $y''(x)=\tfrac{d}{dx} w(y(x))=\tfrac{d}{dy} w(y) y'(x)=w'(y) w(y)$ and thus we have to solve the alternate problem
$$yw(y)+w'(y)w(y)=0$$
or in other words
$$\underbrace{w(y)}_{(1)} (\underbrace{y+w'(y)}_{(2)})=0.$$
Solve $(1)=0$ and $(2)=0$ and get $w(y)=0$ and $w(y)=-0.5y^2+C$. Plugging this back into $w(y)=y'$ yields $y'=0$ or $y'=-0.5y^2+C$. The first yields $y=\text{const}$ and thus is no solution as it doesn't fulfills your initial condition.
Now just solve $y'(x)=-0.5y^2(x)+C$. Can you take it from here? E.g. use separation of variables.
Note: You would get the same if you use the comment by John Wayland Bales which is also more elegant.
A: $$y''+yy'=0 \\\times 2\\2y''+2yy'=0\\(2y'+y^2)'=0\\2y'+y^2=const\\2y'+y^2=c_1\\y'=c_1-\frac12 y^2\\put \space x=0\\y'(0)=c_1-\frac12 y(0)^2\\\frac{-1}{2}=c_1-\frac12\\c_1=0\\y'=y^2\\\frac{dy}{dx}=-y^2\\\frac{dy}{-y^2}=dx
$$apply integral to both sides
$$\int_{0}^{x}\frac{dy}{-y^2}=\int_{0}^{x}1ds\\\frac{1}{y}-\frac{1}{y(0)}=x\\
\frac{1}{y}-1=x$$
A: In order to take advantage of the hint, you need to take the suggested equality and plug it into the ODE. You have the right idea ($w' + yw = 0$), but there is an error in your work. You need to apply the chain rule to the left side of $w(y(x)) = y'(x)$. Additionally, there will be a substitution required once the chain rule is completed successfully. From there, you should end up with a 1st order ODE that can be solved using separation of variables. Keep in mind that you eliminate the trivial solution $f = 0$ if you divide both sides of an equation by f (where f is the independent variable).
