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The differential equation of the family of curves $y = Ae^{3x} + Be^{-3x}$, where A and B are arbitrary constants, is |
$\frac{d^2y}{dx^2}+9y=0$ $\frac{d^2y}{dx^2}+3y=0$ $\frac{d^2y}{dx^2}=9y$ $\frac{d^2y}{dx^2}=3y$ |
$\frac{d^2y}{dx^2}=9y$ |
The correct answer is Option (3) → $\frac{d^2y}{dx^2}=9y$ ** Given family: $y= A e^{3x}+B e^{-3x}$ First derivative: $y'=3A e^{3x}-3B e^{-3x}$ Second derivative: $y''=9A e^{3x}+9B e^{-3x}$ Since $9A e^{3x}+9B e^{-3x}=9y$, the differential equation is $y''-9y=0$ The differential equation of the family is $y''-9y=0$. |
