# Solving Nonlinear ODEs I

#### hatguy

##### New member
I need to solve 2 ODEs:

1. 2. Last edited by a moderator:

#### Ackbach

##### Indicium Physicus
Staff member
Regarding 1:

If you let $u=\ln(y)$, then the equation
$$\ln(y)=\left( \frac{y'}{y}\right)^{\!\!2}+x\,\frac{y'}{y}$$
reduces to
$$u=(u')^{2}+xu'.$$
If you view this is a quadratic in $u'$, you can find that
$$u'=\frac{-x \pm \sqrt{x^{2}+4u}}{2}.$$
Not sure where you could go from here. You could try to make it exact.

Regarding 2:

The substitution $u=y^{2}$ renders the equation Ricatti. Have fun with that!

#### Ackbach

##### Indicium Physicus
Staff member
Further update on Number 1: the additional substitution $v=-1+\sqrt{1+4u}$ renders the equation separable, I think.

#### Jester

##### Well-known member
MHB Math Helper
Q1 Differentiate what Ackbach has giving $u''(2u'+x) = 0$ - two cases to consider.

Q2 As Ackbach said let $u = y^2$, further let $t = x^4$. Your new equation should be homogeneous.

Last edited by a moderator:

#### hatguy

##### New member
Q1 Differentiate what Ackbach has giving $u''(2u'+x) = 0$ - two cases to consider.

Q2 As Ackbach said let $u = y^2$, further let $t = x^4$. Your new equation should be homogeneous.
Thanks, everyone, I found my solutions. And yes, a left bracket shouldn't be where it is now.