Second order differential equations double substitution The question is $$x\frac{d^2y}{dx^2} + x\left(\frac{dy}{dx}\right)^2 + 0.5\frac{dy}{dx}=0.25$$
Hint: Use $x=t^2$ , $u=e^y$, then derive a differential equation for $u(t)$
I used the $x=2t$ and differentiated it to get $\frac{dx}{dt}=2t$  and $\frac{dy}{dx}=\frac{dy}{dt}\cdot \frac{1}{2t}$, Differentiating again $\frac{d^2y}{dx^2}=0.25(t^{-2})\frac{d^2y}{dt^2}-0.25(t^{-3})\frac{dy}{dt}$
Subbing again, I got $\frac{d^2y}{dt^2} +(\frac{dy}{dt})^2=1$ 
Am I going about the question the right way?
 A: (BIG) HINT:
$$x\cdot\frac{\text{d}^2y}{\text{d}x^2}+x\left(\frac{\text{d}y}{\text{d}x}\right)^2+0.5\cdot\frac{\text{d}y}{\text{d}x}=0.25\Longleftrightarrow$$
$$xy''(x)+xy'(x)^2+\frac{y'(x)}{2}=\frac{1}{4}\Longleftrightarrow$$

Let $y'(x)=v(x)$, which gives $y''(x)=v'(x)$:
Solve Riccati's equation $xv(x)^2+\frac{v(x)}{2}+xv'(x)=\frac{1}{4}$:
Subtract $xv(x)^2+\frac{v(x)}{2}$ from both sides:

$$xv'(x)=-xv(x)^2-\frac{v(x)}{2}+\frac{1}{4}\Longleftrightarrow$$
$$v'(x)=-v(x)^2-\frac{v(x)}{2x}+\frac{1}{4x}\Longleftrightarrow$$

Let $v(x)=-u(x)$, which gives $v'(x)=-u'(x)$:

$$-u'(x)=-u(x)^2+\frac{u(x)}{2x}+\frac{1}{4x}\Longleftrightarrow$$
$$u'(x)=u(x)^2-\frac{u(x)}{2x}-\frac{1}{4x}\Longleftrightarrow$$

Let $u(x)=-\frac{w'(x)}{w(x)}$, which gives $u'(x)=\frac{w'(x)^2}{w(x)^2}-\frac{w''(x)}{w(x)}$:

$$-w''(x)=\frac{w'(x)}{2x}-\frac{w(x)}{4x}\Longleftrightarrow$$
$$w''(x)+\frac{w'(x)}{2x}-\frac{w(x)}{4x}=0\Longleftrightarrow$$

Let $t=i\sqrt{x}$, which gives $x=-t^2$:

$$w''(x)-\frac{w'(x)}{2t^2}+\frac{w(x)}{4t^2}=0\Longleftrightarrow$$

Using the chain rule, $w'(x)=\frac{\text{d}w(t)}{\text{d}t}(w'(t))$:

$$\frac{w''(t)}{4t^2}+\frac{w(t)}{4t^2}=0\Longleftrightarrow$$
$$w''(t)+w(t)=0\Longleftrightarrow$$

Assume a solution will be proportional to $e^{\lambda t}$ for some constant $\lambda$.
Substitute $w(t)=e^{\lambda t}$ into the differential equation:

$$\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)+e^{\lambda t}=0\Longleftrightarrow$$

Substitute $\frac{\text{d}^2}{\text{d}t^2}\left(e^{\lambda t}\right)=\lambda^2e^{\lambda t}$:

$$\lambda^2e^{\lambda t}+e^{\lambda t}=0\Longleftrightarrow$$
$$e^{\lambda t}\left(\lambda^2+1\right)=0\Longleftrightarrow$$

Since $e^{\lambda t}\ne0$ for any finite $\lambda$, the zeros must come from the polynomial:

$$\lambda^2+1=0\Longleftrightarrow$$
$$\lambda=\pm i$$
