Solving ODEs: y=xp+f(p): Show General Solution & Second Solution

[saul goodman]

I've been asked to consider differential equations of the form y=xp+f(p), where p=dy/dx, and to show that the general solution is y=cx+f(c).

Substituting in p, the original equation is y=x(dy/dx)+f(dy/dx), and by differentiating I get:

dy/dx=x(d2y/dx2) + dy/dx + d2y/fx2 * f`(dy/dx).

Substracting dy/dx from both sides:

0 = x(d2y/dx2) + d2y/fx2 * f`(dy/dx).

Factorising d2y/dx2:

0=d2y/dx2(x+f`(dy/dx))

Thus in one case d2y/dx2=0 => dy/dx=c => p=c. Substituting into the original equation we get y=cx+f(c) so that part is done.

It then asks me to further show that there is a second solution obeying the differentiatial equation:

d(f(p))/dp + x = 0

Now I suspected you could solve this by looking at the second case above where x+f`(dy/dx)=0 but I'm not sure how. I noticed you get this new differential equation by simply differentiating the original equation by p. However surely I can't ignore the case of x+f`(dy/dx)=0, it must be involved somehow?

Thanks for any help

 

[lippyka]

you can provide! Yes, you are correct that the second case needs to be taken into account when solving for the second solution. To do this, you need to solve the equation x+f`(dy/dx)=0 for dy/dx, which will give you the value of p, and then you can plug this into the original equation to get y=cx+f(c).