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The substitution $\frac{d y}{d x}=z$, reduces the differential equation $\frac{d^2 y}{d x^2}-\frac{d y}{d x}=0$ to a differential equation whose solution is $\mathrm{z}=$
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$\mathrm{Ae}^{\mathrm{X}}$
$\begin{aligned} & \text { Here, } \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{z} \Rightarrow \frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{dx}^2}=\frac{\mathrm{dz}}{\mathrm{dx}} \\ & \Rightarrow \frac{\mathrm{dz}}{\mathrm{dx}}-\mathrm{z}=0 \Rightarrow \frac{\mathrm{dz}}{\mathrm{dx}}=\mathrm{z} \Rightarrow \frac{\mathrm{dz}}{\mathrm{z}}=\mathrm{dx} \\ & \Rightarrow \log _{\mathrm{e}}|\mathrm{z}|=\mathrm{cx} \\ & \Rightarrow \mathrm{z}=\mathrm{e}^{\mathrm{cx}}=\mathrm{e}^{\mathrm{c}} \cdot \mathrm{e}^{\mathrm{x}} \\ & \Rightarrow \mathrm{z}=\mathrm{A} \cdot \mathrm{e}^{\mathrm{x}} \cdot \mathrm{A}=\mathrm{e}^{\mathrm{c}}\end{aligned}$
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