Question

how to solve for y' in (y')^2 + ty' + 4y=0. Please show the steps

Answer 1

You have an equation that's quadratic in \(y'\). Complete the square:
\[\begin{align*}
(y')^2+ty'+4y&=0\\\\
(y')^2+ty'+\frac{t^2}{4}-\frac{t^2}{4}+4y&=0\\\\
\left(y'+\frac{t}{2}\right)^2-\frac{t^2}{4}+4y&=0\\\\

\end{align*}\]

Answer 2

clear explanation. thank you