Question

Show that y = ae^–3x + b, where a and b are arbitrary constants, is a solution of the differential equation ${\displaystyle \frac{{\text{d}}^{2}y}{\text{d}{x}^2} +3\frac{\text{d}y}{\text{d}x}}$ = 0

Answer 1

Given y = ae^–3x + b  .......(1)

Differentiating equation (1) w.r.t ‘x’, we get

${\displaystyle \frac{\text{d}y}{\text{d}x}={\text{ae}}^{-33x}(-3)+0}$

${\displaystyle \frac{\text{d}y}{\text{d}x}={\text{ae}}^{-3x}(-3)}$  ........${\displaystyle [{\text{ae}}^{-3x}=-\frac{1}{3}\frac{\text{d}y}{\text{d}x}]}$

Again differentiating, we get

${\displaystyle \frac{{\text{d}}^{2}y}{\text{d}{x}^2}={\text{ae}}^{-3x}(+9)}$

${\displaystyle \frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-\frac{1}{3}\frac{\text{d}y}{\text{d}x}\times 9}$

${\displaystyle \frac{{\text{d}}^{2}y}{\text{d}{x}^2}=-3\frac{\text{d}y}{\text{d}x}}$

${\displaystyle \frac{{\text{d}}^{2}y}{\text{d}{x}^2}+3\frac{\text{d}y}{\text{d}x}}$ = 0

Therefore, y = ae^–3x + b is a solution of the given differential equation.