MHT CET · Maths · Differential Equations
The differential equation representing the family of curves \(y^2=2 \mathrm{c}(x+\sqrt{\mathrm{c}})\), where \(\mathrm{c}\) is a positive parameter, is of
- A order 1, degree 4
- B order 2, degree 3
- C order 2, degree 4
- D order 1, degree 3
Answer & Solution
Correct Answer
(D) order 1, degree 3
Step-by-step Solution
Detailed explanation
\(y^2=2 \mathrm{c}(x+\sqrt{\mathrm{c}})... (i)\)
Differentiating w.r.t. \(x\), we get
\(2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{c}... (ii)\)
Substituting (ii) in (i), we get
\(\begin{aligned}
& y^2=2 y \frac{\mathrm{d} y}{\mathrm{~d} x}\left(x+\sqrt{y \frac{\mathrm{d} y}{\mathrm{~d} x}}\right) \\
& \Rightarrow y=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 \frac{\mathrm{d} y}{\mathrm{~d} x} \sqrt{y \frac{\mathrm{d} y}{\mathrm{~d} x}} \\
& \Rightarrow\left(y-2 x \frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2=4 y\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3
\end{aligned}\)
This is a differential equation of order 1 and degree 3 .
Differentiating w.r.t. \(x\), we get
\(2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=2 \mathrm{c}... (ii)\)
Substituting (ii) in (i), we get
\(\begin{aligned}
& y^2=2 y \frac{\mathrm{d} y}{\mathrm{~d} x}\left(x+\sqrt{y \frac{\mathrm{d} y}{\mathrm{~d} x}}\right) \\
& \Rightarrow y=2 x \frac{\mathrm{d} y}{\mathrm{~d} x}+2 \frac{\mathrm{d} y}{\mathrm{~d} x} \sqrt{y \frac{\mathrm{d} y}{\mathrm{~d} x}} \\
& \Rightarrow\left(y-2 x \frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2=4 y\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^3
\end{aligned}\)
This is a differential equation of order 1 and degree 3 .
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