Vtyjkyuk

Given $$ xdfracd^2xdt^2+left (dfracdxdtright )^2=xdfracdxdt,quad x(0)=0, x(1)=1.$$ What is the value of $x(2)$? I tried with $dx/dt=y, dy/dt=(dy/dx)(dx/dt)$ but failed at integration.

Even if this is not a direct answer but it leads to the answer.

This equation can be written as
$$ dfracddt left( xdfracdxdt right)=xdfracdxdt,quad $$
Therfore $ quad xdfracdxdt=ce^t $
Then

$$ int x dx = int ce^t dt$$
so $$frac12x^2(t)=ce^t+d$$
Can you take it from here?

You can divide the equation by $xdotx$ and integrate the equation (chain rule).

$$ fracddotxdotx + fracdotxx = 1 $$

$$ implies ln(dotx) + ln(x) = ln(xdotx) = t+ c $$

$$ implies xdotx = ke^t $$

The last equation is separable again.

Let$p=fracdxdt$ then $fracd^2xdt^2=p fracdpdx$ and the equation becomes
begineqnarray*
fracdpdx=fracx-px
endeqnarray*
or $p=0$. The equation above is easily solved by the substitution $p=xu$ ... good luck.

$$xdfracd^2xdt^2+left (dfracdxdtright )^2=xdfracdxdt,quad x(0)=0, x(1)=1.$$
$$xx''+(x')^2=x'x$$
$$implies (x'x)'=x'x$$
Substitute $z=x'x$
$$z'= z$$
This last equation is separable...