The order of the differential equation y = c_ | vedclass

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Question

(C) Given equation: $y = c_{1} e^{c_{2}+x} + c_{3} e^{c_{4}+x}$ Using the property of exponents $e^{a+b} = e^a \cdot e^b$,we can rewrite the equation

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$1$

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(C) Given equation: $y = c_{1} e^{c_{2}+x} + c_{3} e^{c_{4}+x}$ Using the property of exponents $e^{a+b} = e^a \cdot e^b$,we can rewrite the equation as: $y = c_{1} e^{c_{2}} e^{x} + c_{3} e^{c_{4}} e^{x}$ Factor out $e^{x}$: $y = (c_{1} e^{c_{2}} + c_{3} e^{c_{4}}) e^{x}$ Since $c_{1}, c_{2}, c_{3}, c_{4}$ are constants,the term $(c_{1} e^{c_{2}} + c_{3} e^{c_{4}})$ is also a constant. Let $A = c_{1} e^{c_{2}} + c_{3} e^{c_{4}}$. Then the equation simplifies to: $y = A e^{x}$ Differentiating with respect to $x$: $\frac{dy}{dx} = A e^{x}$ Since $y = A e^{x}$,we have: $\frac{dy}{dx} = y$ This is a first-order differential equation. Therefore,the order is $1$.

The order of the differential equation $y = c_{1} e^{c_{2}+x} + c_{3} e^{c_{4}+x}$ is

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