If [(x + 1)2 f(x) – g(x)] is the particular integral of the differential equation shown below, then

\({\left( {x + 1} \right)^2}\frac{{{d^2}y}}{{d{x^2}}} - 3\left( {x + 1} \right)\frac{{dy}}{{dx}} + 4y = {x^2}\)

1
f(x) = log (x + 1)2 and g(x) = 2x + 5
2
\(f\left( x \right) = \frac{{\log {{\left( {x + 1} \right)}^2}}}{2}\) and \(g\left( x \right) = \frac{{2x + 1}}{3}\)
3
f(x) = log (x + 1) and g(x) = 5x + 1
4
\(f\left( x \right) = \frac{1}{2}{\left[ {\log \left( {x + 1} \right)} \right]^2}\) and  \(g\left( x \right) = \frac{{8x \;+\; 7}}{4}\)

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