- #1
saul goodman
- 6
- 0
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
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