MHT CET · Maths · Differential Equations
The order of the differential equation, whose solution is \(y=\left(C_1+C_2\right) \mathrm{e}^x+C_3 \mathrm{e}^{x+C_4}\), is
- A 4
- B 1
- C 3
- D 2
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
y & =\left(C_1+C_2\right) \mathrm{e}^x+\mathrm{C}_3 \mathrm{e}^{x+\mathrm{C}_4} \\
& =\mathrm{Ae}^x+\mathrm{Be}^x \text {, where } \mathrm{A}=\mathrm{C}_1+\mathrm{C}_2 \text { and } \mathrm{B}=\mathrm{C}_3 \mathrm{e}^{\mathrm{C}_4} \\
& =(\mathrm{A}+\mathrm{B}) \mathrm{e}^x \\
& =\mathrm{De}^x, \text { where } \mathrm{D}=\mathrm{A}+\mathrm{B}
\end{aligned}\)
This equation consist of one arbitrary constant.
y & =\left(C_1+C_2\right) \mathrm{e}^x+\mathrm{C}_3 \mathrm{e}^{x+\mathrm{C}_4} \\
& =\mathrm{Ae}^x+\mathrm{Be}^x \text {, where } \mathrm{A}=\mathrm{C}_1+\mathrm{C}_2 \text { and } \mathrm{B}=\mathrm{C}_3 \mathrm{e}^{\mathrm{C}_4} \\
& =(\mathrm{A}+\mathrm{B}) \mathrm{e}^x \\
& =\mathrm{De}^x, \text { where } \mathrm{D}=\mathrm{A}+\mathrm{B}
\end{aligned}\)
This equation consist of one arbitrary constant.
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