Question

3x(1-x2)y2dy/dx+(2x2-1)y3=ax3​

Answer 1

Answer:

3x(1-x²)y² dy/dx + ( 2x²-1 )y³ = ax³

⇒dy/dx +  [( 2x² − 1 )y³ / [3x * (1−x²) * y²] ] = ax³ / [ 3x * (1−x²) * y² ]

⇒y² dy/dx + [(2x²−1)*y³ / 3x(1−x²)] = ax³ / 3x(1−x²)

Substituting y³ = t so the equation will be

⇒1/3 * [ dt/dx ] + [ ( 2x²−1 )t / 3x(1−x²) ] = ax³ / 3x(1−x²)

after this the integrating, factor is   1 / [x * √1−x²]