MHT CET · Maths · Differential Equations
A particular solution of \(3 \mathrm{e}^x \tan \mathrm{y} \mathrm{d} x+\left(1-\mathrm{e}^x\right) \sec ^2 \mathrm{y} \mathrm{dy}=0\) with \(y(1)=\frac{\pi}{4} \quad\) is
- A \(\tan \mathrm{y}=\left(\frac{1-\mathrm{e}^3}{1-\mathrm{e}^x}\right)^3\)
- B \(\tan \mathrm{y}=\left(\frac{1-\mathrm{e}^2}{1-\mathrm{e}^x}\right)^3\)
- C \(\tan \mathrm{y}=\left(\frac{1-\mathrm{e}}{1-\mathrm{e}^x}\right)^3\)
- D \(\tan \mathrm{y}=\left(\frac{1-\mathrm{e}^x}{1-\mathrm{e}}\right)^3\)
Answer & Solution
Correct Answer
(D) \(\tan \mathrm{y}=\left(\frac{1-\mathrm{e}^x}{1-\mathrm{e}}\right)^3\)
Step-by-step Solution
Detailed explanation
\(\frac{3 \mathrm{e}^x}{1-\mathrm{e}^x} \mathrm{d} x = -\frac{\sec ^2 \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy}\) \(\int \frac{3 \mathrm{e}^x}{1-\mathrm{e}^x} \mathrm{d} x = -\int \frac{\sec ^2 \mathrm{y}}{\tan \mathrm{y}} \mathrm{dy}\)
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