JEE Mains · Maths · STD 12 - 7.2 definite integral
For \(x \in R,x \ne 0\), if \(y(x)\) is a differentiable function such that \(x\int\limits_1^x {y\left( t \right)} dt = \left( {x + 1} \right)\int\limits_1^x {ty\left( t \right)} dt\) , then \(y(x)\) equals (where \(C\) is a constant)
- A \(c{x^3}{e^{\frac{1}{x}}}\)
- B \(\frac{c}{{{x^2}}}{e^{ - \frac{1}{x}}}\)
- C \(\frac{c}{{{x}}}{e^{ - \frac{1}{x}}}\)
- D \(\frac{{c{e^{ - \frac{1}{x}}}}}{{{x^3}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{{c{e^{ - \frac{1}{x}}}}}{{{x^3}}}\)
Step-by-step Solution
Detailed explanation
\(x\int\limits_1^x {y(t)dt = x\int\limits_1^x {ty(t)dt + \int\limits_1^x {ty(t)dt} } } \) Differentiate \(w r:\) to \(x\) \(\int\limits_1^x {y(t)dt + x[y(x) - y(1)]} \) \( = \int\limits_1^x {ty(t)dt + x[xy(x) - y(1)] + xy(x) - y(1)} \)…
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