MHT CET · Maths · Differential Equations
If \(\mathrm{m}\) is order and \(\mathrm{n}\) is degree of the differential equation \(\left(\frac{d^2 y}{d x^2}\right)^5+4 \frac{\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1\), then
- A m=3, n=1
- B m=3, n=2
- C m=3, n=3
- D m=3, n=5
Answer & Solution
Correct Answer
(B) m=3, n=2
Step-by-step Solution
Detailed explanation
\(\text {We have }\left(\frac{d^2 y}{d x^2}\right)^5+\frac{4\left(\frac{d^2 y}{d x^2}\right)}{\left(\frac{d^3 y}{d x^3}\right)}+\left(\frac{d^3 y}{d x^3}\right)=x^2-1\)
\(\therefore\left(\frac{d^3 y}{d x^3}\right)\left(\frac{d^2 y}{d x^2}\right)^5+4\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d^3 y}{d x^3}\right)^2=\) \(\left(x^2-1\right)\left(\frac{d^3 y}{d x^3}\right)\)
\(\therefore \text { order }=3 \text {, degree }=2\)
\(\therefore\left(\frac{d^3 y}{d x^3}\right)\left(\frac{d^2 y}{d x^2}\right)^5+4\left(\frac{d^2 y}{d x^2}\right)\left(\frac{d^3 y}{d x^3}\right)^2=\) \(\left(x^2-1\right)\left(\frac{d^3 y}{d x^3}\right)\)
\(\therefore \text { order }=3 \text {, degree }=2\)
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