Question

The differential equation of the family of curves $y=A{e}^{3x}+B{e}^{5x}$, where $A$ and $B$ are arbitrary constants, is

• A. $\frac{d^{2}y}{d{x}^2}+8\frac{dy}{dx}+15y=0$
• B. $\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}+y=0$
• C. $\frac{d^{2}y}{d{x}^2}-8\frac{dy}{dx}+15y=0$
• D. None of the above

Answer 1

Answer: (C)

$\frac{d^{2}y}{d{x}^2}-8\frac{dy}{dx}+15y=0$

$y=A{e}^{3x}+B{e}^{5x}.......\left(i\right)\frac{dy}{dx}=3A{e}^{3x}+5B{e}^{5x}$

Substituting $B{e}^{5x}$ from $\left(i\right)$

$\frac{dy}{dx}=3A{e}^{3x}+5(y-A{e}^{3x})\frac{dy}{dx}=5y-2A{e}^{3x}.....\left(ii\right)\frac{d^{2}y}{d{x}^2}=5\frac{dy}{dx}-6A{e}^{3x}...\left(iii\right)$

Multiplying $\left(ii\right)$ by $3$

$\Rightarrow 3\frac{dy}{dx}=15y-6A{e}^{3x}$

substituting $-6A{e}^{3x}$ from $\left(iii\right)$

$\Rightarrow 3\frac{dy}{dx}=15y+\frac{d^{2}y}{d{x}^2}-5\frac{dy}{dx}\Rightarrow \frac{d^{2}y}{d{x}^2}-8\frac{dy}{dx}+15y=0$

So option $C$ is correct