MHT CET · Maths · Differential Equations
The differential equation whose solution represents the family \(x^2 y=4 e^x+c\), where \(c\) is an arbitrary constant, is
- A \(x \frac{\mathrm{dy}}{\mathrm{d} x}+x \mathrm{y}=0\)
- B \(x^2 \frac{\mathrm{dy}}{\mathrm{d} x}+(2 xy-4e^x)=0\)
- C \(x \frac{\mathrm{dy}}{\mathrm{d} x}+(x-2) \mathrm{y}=0\)
- D \(x \frac{\mathrm{dy}}{\mathrm{d} x}+(2-x) \mathrm{y}=0\)
Answer & Solution
Correct Answer
(B) \(x^2 \frac{\mathrm{dy}}{\mathrm{d} x}+(2 xy-4e^x)=0\)
Step-by-step Solution
Detailed explanation
\(\frac{d}{dx}(x^2 y) = \frac{d}{dx}(4e^x + c)\) \(2xy + x^2 \frac{dy}{dx} = 4e^x\)
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