The differential equation of $y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right)$ is

The differential equation of $y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right)$ is
  1. $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 y=0$
  2. $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0$
  3. $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0$
  4. $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-\mathrm{e}^x+2 y=0$

Solution

$\begin{aligned} & y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right) \\ & \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right)+\mathrm{e}^x(\mathrm{~b}+2 x) \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=y+\mathrm{e}^x(\mathrm{~b}+2 x) ...(i)\\ & \quad \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x(\mathrm{~b}+2 x)+\mathrm{e}^x(2) \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\mathrm{d} y}{\mathrm{~d} x}-y+2 \mathrm{e}^x ...[From(i)]\\ & \Rightarrow \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0 \end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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