Find $dy/dx$ when $y=2$ given $\dfrac{d^2y}{dx^2}=k \left(\dfrac{dy}{dx}\right)^2$

Question

$$\dfrac{d^2y}{dx^2}=k \left(\dfrac{dy}{dx}\right)^2$$ If $dy/dx=1$ when x=0 and y=1 , what is the value of $dy/dx$ when $y=2$?

Would you need to find the value of k? If yes, how would you find this value?

Answer 1

If we let $u(x)=y'(x)$, then the equation for $u$ is: $$u'(x)=ku^2(x)$$ $$\frac{1}{u^2(x)}u'(x)=k$$ $$-\frac{1}{u(x)}=kx+c$$ And after redefining the constant we get that $$u(x)=\frac{1}{c-kx}$$ So we have that: $$y'(x)=\frac{1}{c-kx}$$ $$y(x)=-\frac{1}{k}\log(c-kx)+d$$ From $y(0)=1$ we have that $$1=-\frac{1}{k}\log(c)+d$$ And from $y'(0)=1$ we have that $$1=\frac{1}{c}$$ So $c=1$ and $d=1$. Now let's set $y=2$ $$2=-\frac{1}{k}\log(1-kx)+1$$ $$-k=\log(1-kx)$$ $$1-kx=e^{-k}$$ So $y'(x)$ when $y=2$: $$y'(x)=\frac{1}{e^{-k}}$$ $$y'(x)=e^k$$

Answer 2

Set $u=e^{-ky}$ then $u''=(-ky''+k^2y'^2)u=0$ and thus $u=ax+b$, $y=-\frac1k\ln(ax+b)$.

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Or another shortcut, as $y$ is assumed to not be constant, change the independent parameter to $y$, set $y'=u(y)$, then $y''=u'(y)y'=u'u$ and the ODE reduces to

$u'=ku$ with $u(1)=1$ and looking for $u(2)$.

This be answered without ever computing the function $y(x)$.