[Year 12 Maths] Differential Equation - I have a question on the degree of ordinary differential equation?

[katsicum]

Hi all, so I am aware that does not have a degree since it involves y' within the square root.

However, I wonder what happens if the "y'" is replaced by x instead, can it have a degree of 1 now? I mean if "y'" is replaced by x then can we just treat the root as a constant since it doesn't involve y?
I am referring to the degree of the following

Thank you so much for helping!

 

[Deleted member 4993]

Hi all, so I am aware that View attachment 19055 does not have a degree since it involves y' within the square root.

However, I wonder what happens if the "y'" is replaced by x instead, can it have a degree of 1 now? I mean if "y'" is replaced by x then can we just treat the root as a constant since it doesn't involve y?
I am referring to the degree of the following View attachment 19056

Thank you so much for helping!

You CANNOT assume y' = \(\displaystyle \frac{dy}{dx}\) = x

Because 'x' is a variable in the equation itself. In doing so, you would be assuming the solution.

However, you can assume:

y' = \(\displaystyle \frac{dy}{dx}\) = u

then

y" = \(\displaystyle \frac{d^2y}{dx^2}\) = u'

Then your DE becomes

u' = \(\displaystyle \sqrt{1+u}\)

 

[HallsofIvy]

IF the equation had been \(\displaystyle y''= \sqrt{1+ x}\) then, yes, the differential equation would have been "first degree". However that has nothing at all to do with the original equation, \(\displaystyle y''= \sqrt{1+ y'}\).

To solve \(\displaystyle y''= \sqrt{1+ y'}\), let u= y' so the equation becomes \(\displaystyle u'= \frac{du}{dx}= \sqrt{1+ u}\) and write it as \(\displaystyle \frac{du}{\sqrt{1+ u}}= dx\). To integrate that let v= 1+ u. Then the equation becomes \(\displaystyle \frac{dv}{\sqrt{v}}= v^{-1/2}dv= dx\). Integrating, \(\displaystyle -\frac{2}{3}v^{-3/2}= x+ C_1\).
\(\displaystyle v= 1+ u= \left[\frac{3}{2}(x+ C_1)\right]^{-2/3}\) so \(\displaystyle u= y'= \left[\frac{3}{2}(x+ C_1)\right]^{-2/3}- 1\) and \(\displaystyle y(x)= -\frac{5}{3}\left[\frac{3}{2}(x+ C_1)\right]^{-5/3}- x+ C_2\).


To solve \(\displaystyle y''= \sqrt{1+ x}= (1+ x)^{1/2}\) integrate to get \(\displaystyle y'= \frac{2}{3}(1+ x)^{3/2}+ C_1\) and then integrate again: \(\displaystyle y(x)= \frac{4}{15}(1+ x)^{5/2}+ C_1x+ C_2\).