MHT CET · Maths · Differential Equations
If \(\mathrm{m}\) is order and \(\mathrm{n}\) is degree of the differential equation \(\mathrm{y}=\frac{\mathrm{dp}}{\mathrm{dx}}+\sqrt{\mathrm{a}^2 \mathrm{p}^2-\mathrm{b}^2}\), where \(\mathrm{p}=\frac{\mathrm{dp}}{\mathrm{dx}}\), then the value of \(\mathrm{m}+\mathrm{n}\) is
- A 2
- B 3
- C 4
- D 5
Answer & Solution
Correct Answer
(C) 4
Step-by-step Solution
Detailed explanation
\(
\begin{aligned}
& y=\frac{d p}{d x}+\sqrt{a^2 p^2-b^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)+\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2} \\
& \therefore y-\frac{d^2 y}{d x^2}=\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2}
\end{aligned}
\)
Squaring both sides, we write
\(
\begin{aligned}
& y^2+\left(\frac{d^2 y}{d x}\right)^2-2 y \frac{d^2 y}{d x^2}=a^2\left(\frac{d y}{d x}\right)^2-b^2 \\
& \therefore \text { order }=2 \text { and degree }=2
\end{aligned}
\)
\begin{aligned}
& y=\frac{d p}{d x}+\sqrt{a^2 p^2-b^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)+\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2} \\
& \therefore y-\frac{d^2 y}{d x^2}=\sqrt{a^2\left(\frac{d y}{d x}\right)^2-b^2}
\end{aligned}
\)
Squaring both sides, we write
\(
\begin{aligned}
& y^2+\left(\frac{d^2 y}{d x}\right)^2-2 y \frac{d^2 y}{d x^2}=a^2\left(\frac{d y}{d x}\right)^2-b^2 \\
& \therefore \text { order }=2 \text { and degree }=2
\end{aligned}
\)
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