Question

solution of this question

Answer 1

Answer will be that  the degree of the given ordinary differential equation is 8.

That is found by successively squaring the given equation four times to remove square root sign completely.

$\sqrt{1+y^2}+\sqrt{1+y'^2}+\sqrt{1+y''^2}=\sqrt{1+x^2}\\\sqrt{1+y^2}+\sqrt{1+y'^2}=-\sqrt{1+y''^2}+\sqrt{1+x^2}\\Squaring... \: 1+y^2+1+y'^2+2\sqrt{(1+y^2)(1+y'^2)}\\=1+y''^2+1+x^2-2\sqrt{(1+y''^2)(1+x^2)}\\\\(y^2+y'^2-x^2-y''^2)^2\\=2^2[(1+y^2)(1+y'^2)+(1+y''^2)(1+x^2) ]\\-8\sqrt{(1+y^2)(1+y'^2)(1+y''^2)(1+x^2)}$

Now if we square the last line of the above set of equations, we get a degree of 8. All square roots are eliminated.  The maximum power or exponent on a term will be 8.

Answer 2

sqrt{1+y^2}+\sqrt{1+y'^2}+\sqrt{1+y''^2}

=\sqrt{1+x^2}\\\sqrt{1+y^2}+\sqrt{1+y'^2}=-

sqrt{1+y''^2}+\sqrt{1+x^2}\\Squaring... \:

1+y^2+1+y'^2+2\sqrt{(1+y^2)(1+y'^2)}

\\=1+y''^2+1+x^2-2\sqrt{(1+y''^2)(1+x^2)}\\\

\(y^2+y'^2-x^2-y''^2)^2\\=2^2[(1+y^2

)(1+y'^2)+(1+y''^2)(1+x^2) ]\\-8\sqrt{(1+y^2)(1+y'^2

)(1+y''^2)(1+x^2)}\end{}​√​1+y​2​​​​​+√​1+y​′2​​​​​+√​1+y​′′2​​​​​=√

​1+x​2​​​​​​√​1+y​2​​​​​+√​1+y​′2​​​​​=−√​1+y​′′2​​​​​+√​1+x​2​​​​​​1+y​2​​+1+y​′2​

​+2√​(1+y​2​​)(1+y​′2​​)​​​​=1+y​′′2​​+1+x​2​​−2√​(1+y​′′2​​)(1+x​2​​)​​​​​(y​2​

​+y​′2​​−x​2​​−y​′′2​​)​2​​​=2​2​​[(1+y​2​​)(1+y​′2​​)+(1+y​′′2​​)(1+x​2​​


​−8√​(1+y​2​​)(1+y​′2​​)(1+y​′′2​​)(1+x​2​​)​​​​​