MHT CET · Maths · Differential Equations
A particular solution of \(\frac{\mathrm{dy}}{\mathrm{d} x}=(x+9 \mathrm{y})^2\), when \(x=0, \mathrm{y}=\frac{1}{27}\) is
- A \(3 x+27 y=\tan \left[3\left(x+\frac{\pi}{12}\right)\right]\)
- B \(3 x+27 y=\tan \left(x+\frac{\pi}{4}\right)\)
- C \(3 x+27 y=\tan \left(x+\frac{\pi}{12}\right)\)
- D \(3 x+27 y=\tan \left[3\left(x+\frac{\pi}{4}\right)\right]\)
Answer & Solution
Correct Answer
(A) \(3 x+27 y=\tan \left[3\left(x+\frac{\pi}{12}\right)\right]\)
Step-by-step Solution
Detailed explanation
Let \(v=x+9y\). Then \(\frac{\mathrm{dv}}{\mathrm{d}x} = 1+9\frac{\mathrm{dy}}{\mathrm{d}x}\).
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