The differential equation of the family of cu | vedclass

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(B) Given the equation $y = a e^{2x} + b e^{5x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = 2a e^{2x} + 5b e^{5x}$  (Equation $1$)Next,d

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$\frac{d^2 y}{d x^2} - 7 \frac{d y}{d x} + 10 y = 0$

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(B) Given the equation $y = a e^{2x} + b e^{5x}$.First,differentiate with respect to $x$:$\frac{dy}{dx} = 2a e^{2x} + 5b e^{5x}$  (Equation $1$)Next,differentiate again with respect to $x$:$\frac{d^2y}{dx^2} = 4a e^{2x} + 25b e^{5x}$  (Equation $2$)We want to eliminate $a$ and $b$. From Equation $1$,multiply by $5$:$5 \frac{dy}{dx} = 10a e^{2x} + 25b e^{5x}$  (Equation $3$)Subtract Equation $2$ from Equation $3$:$5 \frac{dy}{dx} - \frac{d^2y}{dx^2} = 6a e^{2x} \implies a e^{2x} = \frac{5 \frac{dy}{dx} - \frac{d^2y}{dx^2}}{6}$.Alternatively,consider the characteristic equation approach. The roots are $m_1 = 2$ and $m_2 = 5$.The characteristic equation is $(m - 2)(m - 5) = 0$.$m^2 - 7m + 10 = 0$.Replacing $m^k$ with $\frac{d^k y}{dx^k}$,we get $\frac{d^2 y}{dx^2} - 7 \frac{dy}{dx} + 10 y = 0$.

The differential equation of the family of curves given by $y = a e^{2x} + b e^{5x}$,where $a$ and $b$ are parameters,is:

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