Q54 / 110 · VITEEE 2024 · Mathematics
The differential equation corresponding to the family of curves $y=e^x(a x+b)$ is
- A$\frac{d^2 y}{d x^2}+z \frac{d y}{d x}-y=0$
- Correct answerB$\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y=0$
- C$\frac{d^2 y}{d x^2}+2 \frac{d y}{d x}+y=0$
- D$\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}-y=0$
Solution
$\begin{aligned}
& \text { (b) } y=e^x(a x+b) \\
& \Rightarrow y \cdot e^{-x}=a x+b
\end{aligned}$
Differentiating both sides, we get
$e^{-x}\left(\frac{d y}{d x}\right)-y e^{-x}=a$
Again differentiating both sides
$\begin{array}{ll}
e^{-x}\left(\frac{d^2 y}{d x^2}\right)-e^{-x}\left(\frac{d y}{d x}\right)+y e^{-x}-e^{-x}\left(\frac{d y}{d x}\right)=0 \\
\Rightarrow & e^{-x}\left(\frac{d^2 y}{d x^2}-2 \frac{d y}{d x}+y\right)=0 \\
\Rightarrow & \frac{d^2 y}{d x^2}-2\left(\frac{d y}{d x}\right)+y=0
\end{array}$
Hence, the correct answer is option: 'B'.