Question

from a differential equations for y=(a+bx).e power 3x

Answer 1

$y = (a + bx) {e}^{3x} \\ (y \div {e}^{3x} ) = a + bx \\ differentiate \\ (dy \div dx) {e}^{ - 3x} + y( - 3 {e}^{ - 3x} ) = 0 + b \\ diff \\ ( {d}^{2} y \div d {x}^{2} )( {e}^{ - 3x} ) + (dy \div dx)( - 3 {e}^{ - 3x} ) \\ + (dy \div dx)( - 3 {e}^{ - 3x} ) + 9y {e}^{ - 3x} = 0 \\ = ( {e}^{ - 3x}) (( {d}^{2} y \div d {x}^{2} ) + ( - 6)(dy \div dx) + 9y) = 0 \\ = ( {d}^{2} y \div d {x}^{2} ) - 6(dy \div dx) + 9y = 0$
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