Find the differential equation whose general solution is y =…

Find the differential equation whose general solution is $y = c_1(x - c_2)^2$.

Step 1: Differentiate the family

The general solution contains two arbitrary constants $c_1$ and $c_2$. Differentiating $y = c_1(x-c_2)^2$ twice gives $y' = 2c_1(x - c_2)$, $y'' = 2c_1$.

Step 2: Express the constants in terms of $y$, $y'$, $y''$

From the second derivative, $c_1 = \dfrac{y''}{2}$. From the first derivative, $x - c_2 = \frac{y'}{2c_1} = \frac{y'}{y''}.$

Step 3: Substitute back into the original equation

Substituting into $y = c_1(x-c_2)^2$ yields

\[y = \frac{y''}{2}\cdot\left(\frac{y'}{y''}\right)^2 = \frac{y''}{2}\cdot\frac{(y')^2}{(y'')^2} = \frac{(y')^2}{2y''}.\]

Step 4: The differential equation

Rearranging gives

\[2y\,y'' = (y')^2.\]

Step 5: Verification

With $y = c_1(x-c_2)^2$, $y' = 2c_1(x-c_2)$, and $y'' = 2c_1$, $2y\,y'' = 2\cdot c_1(x-c_2)^2 \cdot 2c_1 = 4c_1^2(x-c_2)^2$, $(y')^2 = \bigl[2c_1(x-c_2)\bigr]^2 = 4c_1^2(x-c_2)^2$.

Both sides are equal, so $y = c_1(x-c_2)^2$ is the general solution of $2y\,y'' = (y')^2$. The trivial solution $y \equiv 0$ (corresponding to $c_1 = 0$) also satisfies the equation. For $c_1 \neq 0$ the solution represents a family of parabolas with vertex at $(c_2, 0)$.