Question

If $y=P{e}^{ax}+Q{e}^{bx}$, then prove that $\frac{d^{2}y}{d{x}^2}-(a+b)\frac{dy}{dx}+aby=0$

Answer 1

We have $y=P{e}^{ax}+Q{e}^{bx}$ [On dividing both sides by $e^{bx}$

$\Rightarrow e^{-bx}y=P{e}^{(a-b)x}+Q)$ [On diff. w.r.t. x both sides

$\Rightarrow e^{-bx}{y}^{\prime }-by{e}^{-bx}=P(a-b)e^{(a-b)x}\Rightarrow e^{-bx}(y^{\prime }-by)=P(a-b)e^{(a-b)x}$

On multiplying both the sides by $e^{bx-ax}$, we get:

$\Rightarrow e^{-ax}(y^{\prime }-by)=P(a-b)$ [Againg diff. w.r.t . x both sides

$\Rightarrow e^{-ax}(y"-b{y}^{\prime })-a{e}^{-ax}(y^{\prime }-by)=0\Rightarrow e^{-ax}[y"-b{y}^{\prime }-a{y}^{\prime }+aby]=0$

i.e., $y"-(a+b)"y^{\prime }+aby=0$ or $\frac{d^{2}y}{d{x}^2}-(a+b)\frac{dy}{dx}+aby=0$