Simplify and factor (y')^2 - 4x^2 y' + 4x^4 - x^2, then use the factorisation to solve the differential equation obtained by setting it equal to zero.

Step-by-step solution

1. 1

   Treat $y'$ as the variable and read off a perfect square. Grouping the first three terms,

   $(y')^2-4x^2y'+4x^4=(y'-2x^2)^2$

   because $2\cdot y'\cdot 2x^2=4x^2y'$ and $(2x^2)^2=4x^4$. The leftover term is just $-x^2$.

2. 2

   Recognise the difference of squares. The expression is now

   $(y'-2x^2)^2-x^2=\bigl(y'-2x^2-x\bigr)\bigl(y'-2x^2+x\bigr)$

   using $A^2-B^2=(A-B)(A+B)$ with $A=y'-2x^2$ and $B=x$.

3. 3

   Verify the factorisation by expanding. Multiplying out, $(y'-2x^2)^2-x^2=(y')^2-4x^2y'+4x^4-x^2$, which is the original expression. A numerical spot check with $y'=1.4$, $x=0.7$ gives $-1.5906$ on both sides.

4. 4

   Set each factor to zero to split the differential equation. A product is zero when a factor is, so the single quadratic ODE separates into two simple ones:

   $y'=2x^2+x\qquad\text{or}\qquad y'=2x^2-x$

5. 5

   Integrate each branch.

   $y=\frac{2x^3}{3}+\frac{x^2}{2}+C\qquad\text{or}\qquad y=\frac{2x^3}{3}-\frac{x^2}{2}+C$

   Two one-parameter families, as expected: the equation is quadratic in $y'$, so at each point there are two admissible slopes.

6. 6

   Check one solution back in the original equation. For $y=\tfrac{2x^3}{3}+\tfrac{x^2}{2}$ we have $y'=2x^2+x$, so $y'-2x^2=x$ and

   $(y'-2x^2)^2-x^2=x^2-x^2=0\ \checkmark$

   The same argument with $y'-2x^2=-x$ handles the other family.