Question

solve x^2y''-3xy'+4y=1+x^2

Answer 1

$\texttt{given \: equation : }\tiny \\ \\ {x}^{2}y - 3xy + 4y = 1 + {x}^{2} \\ \\ {x}^{2} y - 3xy = 1 + {x}^{2} - 4y \\ \\ y( {x}^{2} - 3x) = 1 + {x}^{2} - 4y \\ \\ {x}^{2} - 3x = \frac{1 + {x }^{2} - 4y}{y} \\ \\ - 3x = - 3 \\ \\ x = \frac{ - 3}{ - 3} \\ \\ \texttt{x = 1}$

Answer 2

Answer:

solve x^2y''-3xy'+4y=1+x^2