1. ## 2yy''-[(y')^2]+1=0

2yy''-[(y')^2]+1=0

[(x+c2)c1/4]-1/c1 ~?

2. The way to 'attack' this type of second order DE is explained here...

http://www.mathhelpforum.com/math-he...iliar-ode.html

In particular the formula to be used is...

$y^{''} = y^{'}\cdot \frac{dy^{'}}{dy}$ (1)

Kind regards

$\chi$ $\sigma$

3. what is the answer ~?

4. Originally Posted by sainimu78
2yy''-[(y')^2]+1=0

[(x+c2)c1/4]-1/c1 ~?
Use the chain rule and the substituion

$u=y' \implies y''=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u\fra c{du}{dy}$

This gives the first order ODE

$2yu\frac{du}{dy}-u^2+1=0 \iff \frac{du}{dy}-\frac{1}{2y}u=\frac{1}{2y}$

This can be solved via an integrating factor

5. Originally Posted by TheEmptySet
Use the chain rule and the substituion

$u=y' \implies y''=\frac{du}{dx}=\frac{du}{dy}\frac{dy}{dx}=u\fra c{du}{dy}$

This gives the first order ODE

$2yu\frac{du}{dy}-u^2+1=0 \iff \frac{du}{dy}-\frac{1}{2y}u=\frac{1}{2y}$

This can be solved via an integrating factor

I solved this equation two days ,

[(x+c2)c1/4]-1/c1 is different from the key at the textbook

6. The second order DE is...

$2\cdot y\cdot y^{''} - y^{' 2} + 1 = 0$ (1)

... that with the substitution...

$y^{''} = y^{'}\cdot \frac{dy^{'}}{dy}$ (2)

... becomes...

$2\cdot y\cdot y^{'}\cdot \frac{dy^{'}}{dy} = y^{' 2} -1 \rightarrow \frac{y^{'}}{y^{' 2} -1}\cdot dy^{'} = \frac{dy}{2y} \rightarrow$

$\rightarrow \ln (y^{' 2} -1) = \ln y + \ln c_{1} \rightarrow y^{' 2} = 1+ c_{1}\cdot y \rightarrow$

$\rightarrow y^{'} = \pm \sqrt {1 + c_{1}\cdot y}$ (3)

The first order DE we have obtained has separable variables so that it can be solved in 'standard form'...

$\frac{dy}{\sqrt {1 + c_{1}\cdot y}} = \pm dx \rightarrow \frac{2}{c_{1}}\cdot \sqrt {1 + c_{1}\cdot y } = \pm x + c_{2} \rightarrow$

$\rightarrow \frac{4}{c_{1}}\cdot y = x^{2} \pm 2\cdot c_{2}\cdot x + c_{2}^{2} - \frac{4}{c_{1}^{2}} \rightarrow \dots$ (4)

Kind regards

$\chi$ $\sigma$

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# solve 1 y'^2 2yy"=0

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