The differential equation of the family of cu | vedclass

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(B) Given the family of curves: $y = A{e^{3x}} + B{e^{5x}}$Step $1$: Differentiate with respect to $x$:$\frac{{dy}}{{dx}} = 3A{e^{3x}} + 5B{e^{5x}}$St

Vedclass · Accepted Answer

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

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(B) Given the family of curves: $y = A{e^{3x}} + B{e^{5x}}$Step $1$: Differentiate with respect to $x$:$\frac{{dy}}{{dx}} = 3A{e^{3x}} + 5B{e^{5x}}$Step $2$: Differentiate again with respect to $x$:$\frac{{{d^2}y}}{{d{x^2}}} = 9A{e^{3x}} + 25B{e^{5x}}$Step $3$: We eliminate the arbitrary constants $A$ and $B$. The characteristic equation for the roots $m_1 = 3$ and $m_2 = 5$ is $(m - 3)(m - 5) = 0$,which simplifies to $m^2 - 8m + 15 = 0$.Step $4$: Replacing $m^k$ with $\frac{{d^k}y}{{dx^k}}$,we get the differential equation:$\frac{{{d^2}y}}{{d{x^2}}} - 8\frac{{dy}}{{dx}} + 15y = 0$Thus,the correct option is $B$.

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

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