MHT CET · Maths · Differential Equations
The solution of the equation \(x^2 \mathrm{y}-x^3 \frac{\mathrm{dy}}{\mathrm{d} x}=\mathrm{y}^4 \cos x\), where \(\mathrm{y}(0)=1\), is
- A \(\mathrm{y}^3=3 x^2 \sin x\)
- B \(x^3=3 y^3 \sin x\)
- C \(x^3=y^3 \sin x\)
- D \(\mathrm{y}^3=4 x^3 \sin x\)
Answer & Solution
Correct Answer
(B) \(x^3=3 y^3 \sin x\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\mathrm{y}^4} \frac{\mathrm{dy}}{\mathrm{d} x} - \frac{1}{x \mathrm{y}^3} = -\frac{\cos x}{x^3}\) Let \(v=\mathrm{y}^{-3}\), so \(\frac{\mathrm{dv}}{\mathrm{d} x} = -3 \mathrm{y}^{-4} \frac{\mathrm{dy}}{\mathrm{d} x}\): \(\frac{\mathrm{dv}}{\mathrm{d} x} + \frac{3}{x} v = \frac{3 \cos x}{x^3}\)
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