The degree of the differential equation y = a | vedclass

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(D) Given equation: $y = a(1 - e^{-x/a})$$y = a - ae^{-x/a}$$y - a = -ae^{-x/a}$Divide by $-a$: $\frac{a - y}{a} = e^{-x/a}$Taking natural logarithm o

Vedclass · Accepted Answer

Not defined

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(D) Given equation: $y = a(1 - e^{-x/a})$$y = a - ae^{-x/a}$$y - a = -ae^{-x/a}$Divide by $-a$: $\frac{a - y}{a} = e^{-x/a}$Taking natural logarithm on both sides: $\ln(\frac{a - y}{a}) = -\frac{x}{a}$Since $a$ is a parameter,we differentiate with respect to $x$ to eliminate it.However,the equation contains an exponential term involving the parameter $a$ in the exponent,which cannot be eliminated to form a polynomial differential equation in terms of $y'$ and $y''$.Specifically,the presence of the parameter $a$ in the exponent $e^{-x/a}$ makes it impossible to express the differential equation as a polynomial in derivatives.Therefore,the degree of this differential equation is not defined.

The degree of the differential equation $y = a(1 - e^{-x/a})$,where $a$ is a parameter,is:

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