MHT CET · Maths · Differential Equations
The differential equation of \(y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right)\) is
- A \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 y=0\)
- B \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}+y=0\)
- C \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0\)
- D \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-\mathrm{e}^x+2 y=0\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& y=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right) \\
& \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{e}^x\left(\mathrm{a}+\mathrm{b} x+x^2\right)+\mathrm{e}^x(\mathrm{~b}+2 x) \\
& \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=y+\mathrm{e}^x(\mathrm{~b}+2 x) ...(i)\\
& \quad \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{e}^x(\mathrm{~b}+2 x)+\mathrm{e}^x(2) \\
& \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\mathrm{d} y}{\mathrm{~d} x}-y+2 \mathrm{e}^x ...[From(i)]\\
& \Rightarrow \frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}-2 \frac{\mathrm{~d} y}{\mathrm{~d} x}-2 \mathrm{e}^x+y=0
\end{aligned}\)
& 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}\)
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