Differential Equations — Class 12 Maths Solution

.: We have, $y = a{e^{3x}} + b{e^{ - 2x}}$

…(1)
Differentiating (1) w.r.t. $x$,

we get
${y_1} = a{e^{3x}} \cdot 3 + b{e^{ - 2x}}( - 2)$

…(2)
Multiplying (1) by 3 and subtracting it from (2),

we get
${y_1} - 3y = - 5b{e^{ - 2x}}$

…(3)
Again differentiating (3) w.r.t. $x$,

we get
${y_2} - 3{y_1} = - 5b{e^{ - 2x}}( - 2)$
…(4)
Multiplying (3) by 2 and adding it to (4),

we get
${y_2} - 3{y_1} + 2({y_1} - 3y) = 0$
$\Rightarrow {y_2} - {y_1} - 6y = 0$,

which is the required differential equation.