# My alternative way to solve first order higher degree ODE

When solving first order higher degree ODE for example : $$y=-xy'+y'^2+\frac{x^2}{2}$$ I solved it for $$~y~$$ as explained in this link :http://www.solitaryroad.com/c652.html

My problem is NOT how to solve the DE , I know how to solve it and I show below how I solved it. ( My question is shown after the solution , my question is can we solve it in an alternative way ? the alternative way is shown after the solution )

The right solution ( as in the link) :

I called $$~y'~$$ as $$~p~$$ and differentiated the DE w.r.t $$~x~$$ which gives me :

$$0=(-2p+x)(1-\frac{dp}{dx})$$ so the general solution is obtained from solving the DE obtained from the second bracket : $$p=x+c$$ substitute in the DE by $$~p~$$ to get the general solution which is : $$y=-x(x+c)+(x+c)^2+\frac{x^2}{2}$$ The first bracket gives the singular solution : $$p=\frac{x}{2}$$ substituting by $$~p~$$ in the DE gives the singular solution which is : $$y=\frac{x^2}{4}$$

The alternative solution : why in the above solution , we did not use the fact that $$p = \frac{dy}{dx}$$ for example in the singular solution after obtaining $$p=\frac{x}{2}$$ why we did not complete as following : $$\frac{dy}{dx}= \frac{x}{2}$$ $$y=\frac{x^2}{4}+k$$ The same for the second bracket containing the differentiation of $$~p~$$ , why we did not complete as following : $$\frac{dy}{dx}=x+c$$ $$y=\frac{x^2}{2}+cx+c_1$$

• There is a typo in your question. The ODE is not $y=-xy+y'^2+\frac{x^2}{2}$ , but is $y=-xy'+y'^2+\frac{x^2}{2}$. – JJacquelin Oct 23 '17 at 7:01
• Where's the link? – Dylan Oct 23 '17 at 19:21
• @Dylan I added the link . I am sorry . – MCS Oct 23 '17 at 20:05

$$y=-xy'+y'^2+\frac{x^2}{2}$$ It is false to say that $\quad y=\frac{x^2}{2}+cx\quad$ is solution of the ODE.

Putting $y=\frac{x^2}{2}+cx$ into the ODE leads to :

$\frac{x^2}{2}+cx=-x(x+c)+(x+c)^2+\frac{x^2}{2} \quad\to\quad c^2=0$

Thus $y=\frac{x^2}{2}+cx$ is not solution of the ODE, except the particular case $c=0$.

The correct solving is :

$$p'=\frac{dy}{dx}=x+c \quad\to\quad y=\frac{x^2}{2}+cx+k$$

But, at first begining, the differentiation of the first order ODE transforms it into a second order ODE. In the set of solutions of the second order ODE, of course, the set of solutions of the first order ODE is included, but there are more solutions which are not solutions of the first order ODE. These inadequate solutions must be eliminated.

That is why we always have to bring back the raw result into the initial ODE and check what are the convenient solutions :

$$\frac{x^2}{2}+cx+k=-x(x+c)+(x+c)^2+\frac{x^2}{2} \quad\to\quad k=c^2$$ The correct solution is : $$y=\frac{x^2}{2}+cx+c^2$$

Do the same with $\quad p=\frac{dy}{dx}=\frac{x}{2} \quad\to\quad y=\frac{x^2}{4}+k\quad$ which is correct for the second order ODE, but not for the first order ODE. Putting it back into the first order ODE : $$\frac{x^2}{4}+k=-x(\frac{x}{2} )+(\frac{x}{2})^2+\frac{x^2}{2} \quad\to\quad k=0$$ So, the correct particular solution is $\quad y=\frac{x^2}{4}$

• This is very helpful !.. so this means we have to substitute by p in the DE in all cases to get the solution (whether we will put back p =dy/dx or not) ..So there is no need to put back p =dy/dx .. this is extra step as I understood from you ... One last question , to get the constant that appeared in the singular solution , you substituted by c = -x/2 , how did you get it ? we have to get the envelope , right ? or you computed it using another method ? – MCS Oct 23 '17 at 20:33
• I cannot understand what you mean : "we have to substitute by p in the DE ". Also, I cannot understand whay you mean : " you substituted by c = -x/2 ". : What is substitued by what ? I never wrote c=-x/2 . – JJacquelin Oct 23 '17 at 20:44
• I mean that for example in the last equation you wrote in the post , to know that the value of k=0 , we have to plug p=x/2 in the DE to get y as function of x only as a singular solution and then compare with the solution of the second order ODE to know the value of k .. So putting back p = dy/dx did not let us avoid plugging p=x/2 in the DE , i.e. plugging back p=dy/dx is not an alternative solution as I thought .. In all cases we have to plug p=x/2 in the DE to know the solution . Right ? – MCS Oct 25 '17 at 21:16
• What we have to plug back in the DE is not $p$, what ever $p$ is. What we have to plug back in the DE is $y(x)$ which was obtained and which eventually includes surabondant solutions. If its agree, $y(x)$ is the correct solution. If not, either there is a mistake in the calculus, or the formula representing $y(x)$ contains a parameter which gives correct solutions only for some values, but not for all. In this case, one have to check for what value(s) of this parameter $y(x)$ agrees with the DE. For example in puting $y(x)=\frac{x^2}{4}+k$ into the DE we see that it agrees only when $k=0$. – JJacquelin Oct 26 '17 at 4:32
• This final check is not a general procedure. It is not necessary to do it if we are sure that the calculus that we carried out is without mistake and if we know that the method used don't generate surabondant solutions. We have to proceed to a final checking when the method used is likely to introduce surabondant solutions. For example if the method used transforms the first order DE to a second order DE, it introduces a new arbitrary parameter, thus generating some solutions of the second order DE which are not solutions of the first order DE. – JJacquelin Oct 26 '17 at 4:42