Solve the initial value problem y y'' + 4y^2 - (y')^2 = 0…

Solve the initial value problem $y\,y'' + 4y^2 - (y')^2 = 0, \qquad y(0)=1,\ \ y'(0)=\sqrt{6}$.

Step 1: Reduce the order: $y$ as the independent variable

The equation is autonomous (no explicit $x$), so set $p=y'$ and use $y''=p\,\dfrac{dp}{dy}$ (chain rule). Substituting turns the equation into a first-order ODE relating $p$ to $y$:

\[ y\,p\,\frac{dp}{dy} - p^2 + 4y^2 = 0. \]

Step 2: Substitute $q=p^2=(y')^2$

Since $\dfrac{dq}{dy}=2p\dfrac{dp}{dy}$, i.e.\ $p\dfrac{dp}{dy}=\dfrac12\dfrac{dq}{dy}$, the equation becomes linear in $q(y)$: \[\frac{y}{2}\frac{dq}{dy} - q + 4y^2 = 0 \implies \frac{dq}{dy} - \frac{2}{y}\,q = -8y.\]

Step 3: Solve the linear equation for $q(y)$

Integrating factor: $\mu(y)=\exp\!\left(-\displaystyle\int\frac{2}{y}\,dy\right)=\dfrac{1}{y^2}$. Since $\dfrac{d}{dy}\!\left(\dfrac{q}{y^2}\right)=\dfrac{1}{y^2}\left(\dfrac{dq}{dy}-\dfrac{2q}{y}\right)=\dfrac{1}{y^2}(-8y)=-\dfrac{8}{y}$, integrating gives

\[ \frac{q}{y^2} = -8\ln y + C_1 \implies q = y^2\bigl(C_1-8\ln y\bigr). \]

Step 4: Apply the initial conditions

At $x=0$: $y=1$, $y'=\sqrt6$, so $q(y{=}1)=(\sqrt6)^2=6$. The formula gives $q(1)=1\cdot(C_1-8\ln 1)=C_1$, so $C_1=6$: $q = y^2(6-8\ln y), \qquad\text{i.e.}\qquad (y')^2 = y^2(6-8\ln y).$

Since $y'(0)=\sqrt6>0$, take the positive root: $y'=y\sqrt{6-8\ln y}$.

Step 5: Separate and integrate

Separating variables, $\dfrac{dy}{y\sqrt{6-8\ln y}}=dx$. Substitute $w=6-8\ln y$, $dw=-\dfrac{8}{y}\,dy$:

\[ \int\frac{dy}{y\sqrt{6-8\ln y}} = \int\frac{-dw/8}{\sqrt w} = -\frac{\sqrt w}{4} = -\frac{\sqrt{6-8\ln y}}{4}. \] So $-\dfrac{\sqrt{6-8\ln y}}{4}=x+C_2$; at $x=0,\ y=1$: $-\dfrac{\sqrt6}{4}=C_2$. Hence

\[ \sqrt{6-8\ln y} = \sqrt6-4x. \]

Step 6: Solve for $y$

Squaring: $6-8\ln y = (\sqrt6-4x)^2 = 6-8\sqrt6\,x+16x^2$, so $-8\ln y=-8\sqrt6\,x+16x^2$, i.e.\ $\ln y=\sqrt6\,x-2x^2$:

$y = e^{\sqrt6\,x - 2x^2$.

Step 7: Verification

With $y=e^{\sqrt6x-2x^2}$: $y'=(\sqrt6-4x)y$ and $y''=\bigl[(\sqrt6-4x)^2-4\bigr]y$. Then

\[ yy''+4y^2-(y')^2 = y^2\bigl[(\sqrt6-4x)^2-4\bigr] + 4y^2 - (\sqrt6-4x)^2y^2 = 0.\ \checkmark \]

and $y(0)=e^0=1\ \checkmark$, $y'(0)=(\sqrt6-0)\cdot1=\sqrt6\ \checkmark$.