KCET · Maths · Vector Algebra
The order of the differential equation obtained by eliminating arbitrary constants in the family of curves \(c_{1} y=\left(c_{2}+c_{3}\right) e^{x+c_{4}}\) is
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
We have, \(c_{1} y=\left(c_{2}+c_{3}\right) e^{x+c_{4}}\)
\(\begin{aligned}
&\Rightarrow \quad y=\left(\frac{c_{2}+c_{3}}{c_{1}}\right) e^{c_{4}} \cdot e^{x} \\
&\Rightarrow \quad y=c e^{x} \text { where } c=\frac{c_{2}+c_{3}}{c_{1}} e^{c_{4}}
\end{aligned}\)
Taking log on both sides, we get
\(\begin{aligned}
&\log y=\log \left(c e^{x}\right) \\
&\Rightarrow \quad \log y=\log c+x \log e \\
&\Rightarrow \quad \log y=\log c+x \\
&\text { On differentiating w.r.t.x, we get } \\
&\frac{1}{y} \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=y \\
&\therefore \text { order }=1
\end{aligned}\)
\(\begin{aligned}
&\Rightarrow \quad y=\left(\frac{c_{2}+c_{3}}{c_{1}}\right) e^{c_{4}} \cdot e^{x} \\
&\Rightarrow \quad y=c e^{x} \text { where } c=\frac{c_{2}+c_{3}}{c_{1}} e^{c_{4}}
\end{aligned}\)
Taking log on both sides, we get
\(\begin{aligned}
&\log y=\log \left(c e^{x}\right) \\
&\Rightarrow \quad \log y=\log c+x \log e \\
&\Rightarrow \quad \log y=\log c+x \\
&\text { On differentiating w.r.t.x, we get } \\
&\frac{1}{y} \frac{d y}{d x}=1 \Rightarrow \frac{d y}{d x}=y \\
&\therefore \text { order }=1
\end{aligned}\)
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