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I got a weird result so I'm not sure I did this right

Let the initial condition be $u(0, y_0) = y_0 $ for some $y_0$

By the method of characteristics let

$$\frac{dx}{ds} = 1 \to x = s + A$$ $$x(s=0) = A = 0 \to x(s) = s$$

$$\frac{dy}{ds} = 4x = 4s \to y = 2s^2 + B$$ $$y(s=0) = B = y_0 \to y(s) = 2s^2+y_0$$

$$\frac{du}{ds} = 1 + u^2 \to \arctan u = s + C \to u(s)=\tan(s+C)$$ $$u(0) = \tan(C) = y_0 \to C=\arctan y_0$$

Now we have that $$u(s) = \tan(s+C) = \tan(x + \arctan y_0)$$

Using $$y(s) = 2s^2+y_0 \to y_0 = y - 2s^2$$

We can substitute in $u$ to get $$u(s) = \tan(x + \arctan(y-2s^2)) = \tan(x + \arctan(y-2x^2)) = u(x,y)$$

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Yes, your answer is correct.

$\displaystyle dx=\frac{dy}{4x}=\frac{du}{1+u^2}$

$4x\ dx=dy\implies 2x^2-y=c_1$

$\displaystyle dx=\frac{du}{1+u^2}\implies x-\tan^{-1}u=c_2$

The general solution is given by $f(2x^2-y,x-\tan^{-1}u)=0$

$\implies x-\tan^{-1}u=g(2x^2-y)$

$u(0,y)=y\implies g(-y)=-\tan^{-1}y\implies g(y)=\tan^{-1}y$

The answer is $u=\tan[x-\tan^{-1}(2x^2-y)]$

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