MHT CET · Maths · Differential Equations
The differential equation of family of lines, having \(x\)-intercept as a and \(y\)-intercept as \(b\), is
- A \(\frac{d^2 y}{d x^2}=0\)
- B \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-\frac{\mathrm{d} y}{\mathrm{~d} x}+y=0\)
- C \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x}=y\)
- D \(\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
Answer & Solution
Correct Answer
(A) \(\frac{d^2 y}{d x^2}=0\)
Step-by-step Solution
Detailed explanation
Diff \(\frac{x}{a}+\frac{y}{b}=1\) [two arbitrary constants]
\(\frac{1}{a}+\frac{1}{b} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
Again in diff \(0+\frac{1}{b} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=0\)
\(\Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=0\)
\(\frac{1}{a}+\frac{1}{b} \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=0\)
Again in diff \(0+\frac{1}{b} \cdot \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=0\)
\(\Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=0\)
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