Solve the following differential equations by method (a) or (b) above.

$2y{y}^{\text{'}\text{'}}=y^{\text{'}2}$

The differential equation is $2y{y}^{\text{'}\text{'}}=y^{\text{'}2}$.

A differential equation is made up of one or more functions and their derivatives. The derivatives of a function at a given moment define its rate of change.

The given equation is lacking by the independent variable$x$,thus it can be significantly simplified by using the substitution$y\equiv p$, also implying $y^{\text{'}\text{'}}=\frac{dp}{dx}=\frac{dp}{dy}\frac{dy}{dx}=\frac{dp}{dy}p$by exploiting the chain rule derivative and definition of $p$.

Insert the transformation into the equation,

$2yp\frac{dp}{dy}=p^2$

By factorization, the variable $p$out so as distinct solutions to the equation:

$p\left(2\frac{dp}{dy}-p\right)=0$

It is manifest that one solution to our equation is$p=0$and since $p=\frac{dy}{dx}$, then

$\frac{dy}{dx}=0\Rightarrow y=\text{const}$

With $p=0$, the second case of equation implies

$2y\frac{dp}{dy}-p=0$

The above equation is separated

$2\frac{dp}{p}=\frac{dy}{y}\Rightarrow 2\ln \left|p\right|=\ln \left|y\right|+\hat{A}$

Using the table integral $\int \frac{dx}{x}=\ln \left|x\right|+C$const and labelled the integration constant by $A$. Since $a\ln \left(x\right)=\ln x^a$and $a\ln \left(x\right)=\ln x^a$:

$\ln p^2=\ln \left(\right|y\left|A\right)$

With $A\equiv e^A$and imply

$p^2=A\left|y\right|\Rightarrow p=\sqrt{A\left|y\right|}$

And separate it

$\frac{dy}{\sqrt{\left|y\right|}}=\sqrt{A}dx$

Now solve the equation by using the table integral $\int x^n=\frac{1}{n+1}{x}^{n+1}+$const

$$" width="9">2|y|=xA+B

Here, as an integration constant.

Now, relabel the constants as$\sqrt{A}\to 2A$and$B\to 2B$for convenience and square the previous expression to obtain the second solution for $y$:

$\left|y\right|=(Ax+B{)}^2$

Thus, the solution of the differential equation is $\left|y\right|=(Ax+B{)}^2$.