The substitution frac{dy}{dx}=z reduces the d | vedclass

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(C) Given the substitution $\frac{dy}{dx}=z$,we differentiate with respect to $x$ to get $\frac{d^2y}{dx^2}=\frac{dz}{dx}$.Substituting these into the

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$Ae^{x}$

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(C) Given the substitution $\frac{dy}{dx}=z$,we differentiate with respect to $x$ to get $\frac{d^2y}{dx^2}=\frac{dz}{dx}$.Substituting these into the given differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$,we obtain $\frac{dz}{dx}-z=0$.This is a first-order separable differential equation: $\frac{dz}{dx}=z$.Separating the variables,we have $\frac{dz}{z}=dx$.Integrating both sides,we get $\int \frac{dz}{z} = \int dx$,which results in $\log_{e}|z|=x+C_1$.Taking the exponential of both sides,we get $|z|=e^{x+C_1} = e^{C_1} \cdot e^{x}$.Letting $A = \pm e^{C_1}$,we obtain the solution $z=Ae^{x}$.

The substitution $\frac{dy}{dx}=z$ reduces the differential equation $\frac{d^2y}{dx^2}-\frac{dy}{dx}=0$ to a differential equation whose solution is $z=$

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