यदि [(x + 1)² f(x) – g(x)] नीचे दिखाए गए अवकल समीकरण का विशेष समाकल है, तो

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

माना x + 1 = e^z ताकि z = log (x + 1) और x = e^z – 1

दिया गया समीकरण बन जाता है:

[D(D - 1) – 3D + 4]y = (e^z - 1)², d = d/dz

Or (D² – 4D + 4)y = e^2z – 2e^z + 1

अतिरिक्त समीकरण, m² – 4m + 4 = 0 है

(m – 2)² = 0

∴ \(C.F. = [\left( {{C_1} + {C_2}z} \right){e^{2z}} = \left[ {{C_1} + {C_2}\log \left( {x + 1} \right)} \right]{\left( {x + 1} \right)^2}\)

\(P.I. = \frac{1}{{{{\left( {D - 2} \right)}^2}}}{e^{2z}} - \frac{2}{{{{\left( {D - 2} \right)}^2}}}{e^z} + \frac{1}{{{{\left( {D - 2} \right)}^2}}}0.z\)

= \(\frac{{{z^2}}}{2}{e^2} - \frac{2}{{{{\left( {1 - 2} \right)}^2}}}{e^z} + \frac{1}{{{{\left( {0 - 2} \right)}^2}}}{e^{0.z}}\)

= \(\frac{{{z^2}}}{2}{e^{2z}} - 2{e^z} + \frac{1}{4}\)

= \(\frac{1}{2}{\left( {x + 1} \right)^2}{\left\{ {\log \left( {x + 1} \right)} \right\}^2} - 2\left( {x + 1} \right) + \frac{1}{4}\)

सही उत्तर है:

\(f\left( x \right) = \frac{1}{2}{\left[ {\log \left( {x + 1} \right)} \right]^2}\) और

\(g\left( x \right) = \frac{{8x\; + \;7}}{4}\)