MHT CET · Maths · Differential Equations
If order and degree of the differential equation \(\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\left(\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\right)}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x\), are m and n respectively, then the value of \(\left(m^2+n^2\right)\) is equal to
- A 29
- B 13
- C 5
- D 8
Answer & Solution
Correct Answer
(B) 13
Step-by-step Solution
Detailed explanation
Given differential equation is
\(\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x\)
\(\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5 \cdot \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}+4\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5+\left(\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}\right)^2=\sin x \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\)
Here, order \(=3\) and degree \(=2\)
\(\begin{aligned}
& \therefore \quad m=3, n=2 \\
& \therefore \quad m^2+n^2=3^2+2^2=13
\end{aligned}\)
\(\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5+4 \frac{\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5}{\frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}}+\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=\sin x\)
\(\left(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}\right)^5 \cdot \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}+4\left(\frac{\mathrm{~d}^2 y}{\mathrm{~d} x^2}\right)^5+\left(\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}\right)^2=\sin x \frac{\mathrm{~d}^3 y}{\mathrm{~d} x^3}\)
Here, order \(=3\) and degree \(=2\)
\(\begin{aligned}
& \therefore \quad m=3, n=2 \\
& \therefore \quad m^2+n^2=3^2+2^2=13
\end{aligned}\)
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