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- #2

- Jan 26, 2012

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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!

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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.

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

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Thanks, everyone, I found my solutions. And yes, a left bracket shouldn't be where it is now.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.