# Thread: xyy' = yy' + x(y')^2

1. ## xyy' = yy' + x(y')^2

Hi everyone,
I need help trying to solve the following:

$\displaystyle xyy'' = yy' + x(y')^2$

Mathematica says the solution is $\displaystyle y(x) = c_1e^{x^2}$ but I have no idea how to get there.
I just noticed that the RHS can be written as (xyy')' but made no progress from there.
Any help is appreciated. Thanks!

2. Originally Posted by DJDorianGray
Hi everyone,
I need help trying to solve the following:

$\displaystyle xyy' = yy' + x(y')^2$

Mathematica says the solution is $\displaystyle y(x) = c_1e^{x^2}$ but I have no idea how to get there.
I just noticed that the RHS can be written as (xyy')' but made no progress from there.
Any help is appreciated. Thanks!
Check you ODE - seems like something isn't correct (since y' factors from what you have here).

3. Corrected it, sorry

4. Originally Posted by DJDorianGray
Hi everyone,
I need help trying to solve the following:

$\displaystyle xyy'' = yy' + x(y')^2$

Mathematica says the solution is $\displaystyle y(x) = c_1e^{x^2}$ but I have no idea how to get there.
I just noticed that the RHS can be written as (xyy')' but made no progress from there.
Any help is appreciated. Thanks!
I do not know a special method for solving equations of this type, but your equation can be solved so:

$\displaystyle xyy'' = yy' + x{\left( {y'} \right)^2} \Leftrightarrow \left( {yy'' - {{\left( {y'} \right)}^2}} \right)x = yy' \Leftrightarrow$

$\displaystyle \Leftrightarrow \frac{{y''}}{{y'}} - \frac{{y'}}{y} = \frac{1}{x} \Leftrightarrow {\left( {\ln \left| {y'} \right|} \right)^\prime } - {\left( {\ln \left| y \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow$

$\displaystyle \Leftrightarrow {\left( {\ln \left| {y'} \right| - \ln \left| y \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow {\left( {\ln \left| {\frac{{y'}}{y}} \right|} \right)^\prime } = \frac{1}{x} \Leftrightarrow$

$\displaystyle \Leftrightarrow \ln \left| {\frac{{y'}}{y}} \right| = \ln \left| {{C_1}x} \right| \Leftrightarrow \frac{{y'}}{y} = {C_1}x \Leftrightarrow \int {\frac{{dy}}{y}} = {C_1}\int {x\,dx} \Leftrightarrow$

$\displaystyle \Leftrightarrow \ln y = {C_1}{x^2} + {C_2} \Leftrightarrow y = {e^{{C_1}{x^2} + {C_2}}} = C_2e^{C_1x^2}.$

Do you understand?

5. Try letting $\displaystyle y = e^u$ and see what happens.