Solve the initial value problem y'' + (1)/(y^2)e^(y^2)y'…

Solve the initial value problem $y'' + \frac{1}{y^2}e^{y^2}y' - 2yy'^2 = 0, \qquad y\!\left(-\frac{1}{2e}\right)=1,\quad y'\!\left(-\frac{1}{2e}\right)=e$.

Step 1: The key insight: treat $y'$ as a function of $y$

The equation has no explicit $x$, so set $p=y'$ and treat $p$ as a function of $y$; then $y''=p\,\dfrac{dp}{dy}$. Substituting: $p\,\frac{dp}{dy} + \frac{e^{y^2}}{y^2}\,p - 2yp^2 = 0.$

Step 2: Divide by $p$: a linear equation

Dividing through by $p$ ($p\neq0$): \[\frac{dp}{dy} + \frac{e^{y^2}}{y^2} - 2yp = 0 \implies \frac{dp}{dy} - 2yp = -\frac{e^{y^2}}{y^2},\]

a first-order linear ODE for $p(y)$.

Step 3: Integrating factor

The integrating factor is $\mu(y)=\exp\!\left(-\int 2y\,dy\right) = e^{-y^2}$. Multiplying through: \[\frac{d}{dy}\bigl(e^{-y^2}p\bigr) = -\frac{e^{y^2}}{y^2}\cdot e^{-y^2} = -\frac{1}{y^2}.\]

Integrating, using $-\displaystyle\int \frac{dy}{y^2} = \frac{1}{y}$: \[e^{-y^2}p = \frac{1}{y} + C_1 \implies p = e^{y^2}\left(\frac{1}{y}+C_1\right).\]

Step 4: Apply the initial condition to $p$

At $y=1$, $p=y'=e$. Substituting: \[e = e^{1}\left(\frac{1}{1}+C_1\right) = e(1+C_1) \implies 1 = 1+C_1 \implies C_1 = 0.\] So the relation simplifies to

\[ p = \frac{e^{y^2}}{y}, \qquad \text{i.e.} \qquad y' = \frac{e^{y^2}}{y}. \]

Step 5: Separate and integrate

Separating variables: $y\,e^{-y^2}\,dy = dx.$

With $u=y^2$, $\displaystyle\int y e^{-y^2}\,dy = -\frac{e^{-y^2}}{2}$, so

\[ -\frac{e^{-y^2}}{2} = x + C_2. \]

Step 6: Apply the initial condition to fix $C_2$

At $x=-\dfrac{1}{2e}$, $y=1$: \[-\frac{e^{-1}}{2} = -\frac{1}{2e} + C_2 \implies C_2 = 0.\] So $-\dfrac{e^{-y^2}}{2} = x$, i.e. $e^{-y^2} = -2x$.

Step 7: Solve for $y$

Taking logarithms: $-y^2 = \ln(-2x)$, so $y^2 = -\ln(-2x)$.

$y = \sqrt{-\ln(-2x)$.

valid for $x\in\left(-\tfrac12,\,0\right)$.

Step 8: Verification

At $x=-\dfrac{1}{2e}$: $-2x=\dfrac1e$, so $y^2=-\ln(1/e)=1$, giving $y=1$. $\checkmark$

Differentiating $y^2=-\ln(-2x)$ implicitly: $2yy' = -\dfrac{1}{x}$, so $y' = -\dfrac{1}{2xy}$. At $x=-\dfrac{1}{2e},\,y=1$: $y' = -\dfrac{1}{2\left(-\frac{1}{2e}\right)(1)} = e. $\checkmark$$

Since $e^{y^2}=-\dfrac{1}{2x}$ along the solution, $\dfrac{e^{y^2}}{y} = -\dfrac{1}{2xy}$, which is exactly $y'$ from the line above — confirming the relation $y'=e^{y^2}/y$ from Step 4 holds along the whole curve. Substituting $y'=-\dfrac{1}{2xy}$ and the corresponding $y''$ into the original equation $y''+\dfrac{1}{y^2}e^{y^2}y'-2yy'^2=0$ confirms it is satisfied identically.