# how to solve the DE#2

#### idontknow

$$\displaystyle t^2 s'' +s=ss'^2$$.

#### romsek

Math Team
$s(t) = c_0 \pm t$ works

#### idontknow

$s(t) = c_0 \pm t$ works
I tried $$\displaystyle x=e^z$$ and got $$\displaystyle s''(z)-s'(z) +s(z) =e^{-2z}s(z)s'^2 (z)$$.

#### idontknow

$$\displaystyle s''(z)-s'(z) +s(z) =e^{-2z}s(z)s'^2 (z)$$.

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 ?

#### idontknow

instead of $$\displaystyle p''+p'=pp'+p^2$$ it must be $$\displaystyle p''+p'+p=pp'+p^2 \: \implies p=1$$.
$$\displaystyle p=se^{-z}=1 \implies s=e^{z}=t$$.

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#### idontknow

(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
At what stage did you overlook the solution $s= 0\hspace{1px}$?

#### idontknow

At what stage did you overlook the solution $s= 0\hspace{1px}$?
At all stages .

#### skipjack

Forum Staff
If $s = tv$, $s' = tv' + v$, but you proceeded as though $s' =v' + v$.