how to solve the DE#2

Dec 2015
1,084
169
Earth
\(\displaystyle t^2 s'' +s=ss'^2\).
 
Dec 2015
1,084
169
Earth
\(\displaystyle
s''(z)-s'(z) +s(z) =e^{-2z}s(z)s'^2 (z)

\).

maybe \(\displaystyle s=p(z)e^z \) will reduce the equation into an easier one , \(\displaystyle \; [p'e^z + e^z p ]'-p'e^z =e^{-2z} pe^{z} s'^2 (z)=e^{z}p [p'+p]^{2}\).

\(\displaystyle p'e^z +p''e^z +e^z p =e^{z}p [p'+p] \; \) ; \(\displaystyle \; p''+p'=pp'+p^2 \).

\(\displaystyle [p'+p]'=p[p'+p] \; \) , how to continue from here ?
 
Dec 2015
1,084
169
Earth
instead of \(\displaystyle
p''+p'=pp'+p^2

\) it must be \(\displaystyle
p''+p'+p=pp'+p^2 \: \implies p=1

\).
\(\displaystyle p=se^{-z}=1 \implies s=e^{z}=t\).
 
Last edited:
Dec 2015
1,084
169
Earth
(1) \(\displaystyle t=e^z \) and \(\displaystyle s=e^z v(z)\; \) ; (2) \(\displaystyle \; t=-e^z \) and \(\displaystyle s=e^z v(z)\).

(1) \(\displaystyle e^{2z} \frac{d}{dt}(\frac{de^z v }{de^z }) + e^z v =e^z v (\frac{d e^z v }{de^z } )^2 \; \) ; \(\displaystyle \; e^{2z}\frac{d}{de^z } (e^{-z} \frac{de^z v }{dz} )+e^z v =e^{z} v (v'+v)^2 \).

\(\displaystyle e^z \frac{d}{dz}(v+v')+e^z v =e^z v (v'+v)^2 \implies v''+v'+v=v(v'+v)^2 \).

(2) \(\displaystyle -v''-v'+v=v(v'+v)^2 \) and (1) \(\displaystyle v''+v'+v=v(v'+v)^2 \).

(1)+(2) gives \(\displaystyle v=v(v'+v)^2 \) where \(\displaystyle v_1 =s_1 =0\).

\(\displaystyle v'+v=\pm 1 \; \) ; \(\displaystyle \; (ve^z )'=\pm e^z \; \) ; \(\displaystyle \; v=e^{-z} (c \pm e^z)=ce^{-z}\pm 1\).

\(\displaystyle s=e^z (ce^{-z} \pm 1)=c\pm e^z =c\pm t\).
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
At what stage did you overlook the solution $s= 0\hspace{1px}$?
 

skipjack

Forum Staff
Dec 2006
21,481
2,470
If $s = tv$, $s' = tv' + v$, but you proceeded as though $s' =v' + v$.