MHB Solve the differential equation

[Nikolas7]

help to solve the differential equation
у*у$^{\prime\prime}$=2х*((у$^{\prime})$^2)

 

[Prove It]

help to solve the differential equation
у*у^{\prime\prime}=2х*((у^{\prime})^2)

Put your code in dollar signs to have the LaTeX compile...

Is this your DE?

$\displaystyle \begin{align*} y\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 2\,x\,\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 \end{align*}$

 

[Nikolas7]

Yes, it is true.
What is solution this?

 

[Prove It]

It's messy, but I believe you can separate the variables...

$\displaystyle \begin{align*} y\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 2\,x\,\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 \\ \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= \frac{2\,x}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{1}{t}\,\frac{\mathrm{d}t}{\mathrm{d}x} &= \frac{2\,x}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \textrm{ with } t = \frac{\mathrm{d}y}{\mathrm{d}x} \\ \int{ \frac{1}{t}\,\frac{\mathrm{d}t}{\mathrm{d}x} \,\mathrm{d}x } &= \int{ \frac{2\,x}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \,\mathrm{d}x} \\ \int{ \frac{1}{t}\,\mathrm{d}t} &= \int{ 2\,x\, \frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x} \\ \ln{ \left| t \right| } + C_1 &= \int{ 2\,x\,\frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x} \end{align*}$

For the RHS we use IBP with $\displaystyle \begin{align*} u = 2\,x \implies \mathrm{d}u = 2\,\mathrm{d}x \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v = \frac{1}{y}\,\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x = \frac{1}{y}\,\mathrm{d}y \implies v = \ln{|y|} \end{align*}$ and we have

$\displaystyle \begin{align*} \ln{ \left| t \right| } + C_1 &= 2\,x\ln{ \left| y \right| } - \int{ 2\ln{ \left| y \right| }\,\mathrm{d}x } \\ \ln{ \left| \frac{\mathrm{d}y}{\mathrm{d}x} \right| } + C_1 &= 2\,x\ln{ \left| y \right| } - \int{ 2\ln{ \left| y \right| } \,\mathrm{d}x } \end{align*}$

I am unsure how to continue...

 

[Mathwebster]

If you make the change of variables

$$x = r + 1/2,\;\;\; y = e^s$$

where $$s = s(r)$$, your ODE becomes

$$s'' = 2 r s'^2$$

which now becomes separable. This should help you.

 

[Nikolas7]

Thanks, but I got s"=2r. will try to resolve again.

 

[Nikolas7]

Sorry, you are right: s"=2r(s')^2
because y'=exp(s)s' and y"=exp(s)((s')^2+s")
Many thanks.