Question

Solve xy² (p² + 2) = 2py³ + x³

• A. (y² + x² – c) (y² – x² – cx⁴) = 0
• B. (y² – x² – c) (y² – x² – cx⁴) = 0
• C. (y² – x² – c) (y² – x² + cx⁴) = 0
• D. None of these

Answer 1

Answer: (B)

(y² – x² – c) (y² – x² – cx⁴) = 0

Answer 2

The given equation can be written as ;

(xy²p² – x³) + 2 (xy² – py³) = 0 Ž x(y²p² – x²) + 2y² (x – py) = 0 Ž (py – x) [x(py + x) – 2y²] = 0

If py – x= 0 then y dy – x dx = 0 Ž y² – x² = c₁

If x yp + x² – 2y² = 0 then 2y $\frac{dy}{dx}$ – $\frac{4y^{2}}{x}$ = – 2x Ž $\frac{dt}{dx}$ – $\frac{4}{x}$t = –2x; where t = y²

I . F. = $e^{–\int_{}^{}{\frac{4}{x}dx}}$ = e^{–4 lnx} = $\frac{1}{x^{4}}$

Its solution is t $\left( \frac{1}{x^{4}} \right)$ = $\int_{}^{}{–2x}$· $\frac{1}{x^{4}}$dx

i.e. $\frac{t}{x^{4}}$= $\frac{1}{x^{2}}$ + c₂ i.e., y² = x² + c₂ x⁴

Hence the required solutions is ;

(y² – x² – c) (y² – x² – cx⁴) = 0