MHT CET · Maths · Differential Equations
The order and degree of the differential equation \(y=p x+\sqrt{a^{2} \mathbf{p}^{2}+b^{2}}\), where \(\mathrm{p}=\frac{\mathrm{d} y}{\mathrm{~d} x}\) are respectively
- A \(1, 2\)
- B \(3,1\)
- C 2,1
- D \(1,3\)
Answer & Solution
Correct Answer
(A) \(1, 2\)
Step-by-step Solution
Detailed explanation
Given
\(
\begin{array}{r}
y=p x+\sqrt{a^{2} p^{2}+b^{2}} \\
\therefore y-p x=\sqrt{a^{2} p^{2}+b^{2}}
\end{array}
\)
On squaring both side we get
\(
y^{2}-2 p x y+p^{2} x^{2}=a^{2} p^{2}+b^{2}
\)
\(
\mathrm{y}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2} \mathrm{x}^{2}=\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2}\) \(\quad \ldots\left[\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}, \text { given }\right]
\)
Thus order is 1 and degree is 2 .
\(
\begin{array}{r}
y=p x+\sqrt{a^{2} p^{2}+b^{2}} \\
\therefore y-p x=\sqrt{a^{2} p^{2}+b^{2}}
\end{array}
\)
On squaring both side we get
\(
y^{2}-2 p x y+p^{2} x^{2}=a^{2} p^{2}+b^{2}
\)
\(
\mathrm{y}^{2}-2 \mathrm{xy} \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2} \mathrm{x}^{2}=\mathrm{a}^{2}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{2}+\mathrm{b}^{2}\) \(\quad \ldots\left[\mathrm{p}=\frac{\mathrm{dy}}{\mathrm{dx}}, \text { given }\right]
\)
Thus order is 1 and degree is 2 .
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