MHT CET · Maths · Differential Equations
The differential equation \(\left[\frac{1+\left(\frac{d y}{d x}\right)^2}{\left(\frac{d^2 y}{d x^2}\right)^{\frac{3}{2}}}\right]^2=\mathrm{k} x\) is of
- A order \(=2\), degree \(=3\)
- B order \(=3\), degree \(=2\)
- C order \(=2\), degree \(=2\)
- D order \(=3\), degree \(=3\)
Answer & Solution
Correct Answer
(A) order \(=2\), degree \(=3\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& {\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{k} x} \\
& \therefore:\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^3=(\mathrm{kx})^2 \\
& \therefore \quad\left[1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\right]^3=(\mathrm{k} x)^2\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^3 \\
& \therefore \quad \text { Order }=2, \text { Degree }=3
\end{aligned}\)
& {\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^{\frac{3}{2}}=\mathrm{k} x} \\
& \therefore:\left[\frac{1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2}{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)}\right]^3=(\mathrm{kx})^2 \\
& \therefore \quad\left[1+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\right]^3=(\mathrm{k} x)^2\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^3 \\
& \therefore \quad \text { Order }=2, \text { Degree }=3
\end{aligned}\)
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