4 ( 2 x + 1 ) 2 y " − 4 ( 2 x + 1 ) y ′ + 3 y = ln ( 2 x + 1 ) ) y ( x ) = f ( ln ( 2 x + 1 ) ) y ′ ( x ) = 2 f ′ ( ln ( 2 x + 1 ) ) 2 x + 1 y ′ ′ ( x ) = 4 f ′ ′ ( ln ( 2 x + 1 ) ) ( 2 x + 1 ) 2 − 4 f ′ ( ln ( 2 x + 1 ) ) ( 2 x + 1 ) 2 16 f ′ ′ ( ln ( 2 x + 1 ) ) − 16 f ′ ( ln ( 2 x + 1 ) ) − 8 f ′ ( ln ( 2 x + 1 ) ) + 3 f ( ln ( 2 x + 1 ) ) = ln ( 2 x + 1 ) 16 f ′ ′ ( ln ( 2 x + 1 ) ) − 24 f ′ ( ln ( 2 x + 1 ) ) + 3 f ( ln ( 2 x + 1 ) ) = ln ( 2 x + 1 ) 16 f ′ ′ ( t ) − 24 f ′ ( t ) + 3 f ( t ) = t f ( t ) = c 1 e 3 − 6 4 t + c 2 e 3 + 6 4 t f ( ln ( 2 x + 1 ) ) = c 1 e 3 − 6 4 ln ( 2 x + 1 ) + c 2 e 3 + 6 4 ln ( 2 x + 1 ) = c 1 ( 2 x + 1 ) 3 − 6 4 + c 2 ( 2 x + 1 ) 3 + 6 4 16 f ′ ′ ( t ) − 24 f ′ ( t ) + 3 f ( t ) = t f ( t ) = A t + B f ′ ( t ) = A , f " ( t ) = 0 ⟹ − 24 A + 3 B + 3 A t = t ⟹ 3 A = 1 , A = 1 3 & 3 B = 24 A , B = 8 3 ⟹ f ( t ) = t + 8 3 , f ( ln ( 2 x + 1 ) ) = ln ( 2 x + 1 ) + 8 3 ⟹ y ( x ) = c 1 ( 2 x + 1 ) 3 − 6 4 + c 2 ( 2 x + 1 ) 3 + 6 4 + ln ( 2 x + 1 ) + 8 3 \displaystyle
4(2x+1)^2y" -4(2x+1)y' + 3y = \ln(2x+1))\\
y(x) = f(\ln(2x + 1))\\
y'(x) = 2\frac{f'(\ln(2x + 1))}{2x + 1}\\
y''(x) = \frac{4f''(\ln(2x + 1))}{(2x + 1)^2} -
\frac{4f'(\ln(2x + 1))}{(2x + 1)^2}\\
16f''(\ln(2x + 1)) - 16f'(\ln(2x + 1)) - 8f'(\ln(2x + 1)) + 3f(\ln(2x + 1)) = \ln(2x + 1)\\
16f''(\ln(2x + 1)) - 24f'(\ln(2x + 1)) + 3f(\ln(2x + 1)) = \ln(2x + 1)\\
16f''(t) - 24f'(t) + 3f(t) = t\\
f(t) = c_1e^{\frac{3 - \sqrt{6}}{4}t} + c_2 e^{\frac{3 + \sqrt{6}}{4}t}\\
\begin{aligned}
f(\ln(2x + 1)) &= c_1e^{\frac{3 - \sqrt{6}}{4}\ln(2x + 1)} + c_2 e^{\frac{3 + \sqrt{6}}{4}\ln(2x + 1)}
\\&= c_1 (2x + 1)^{\frac{3 - \sqrt{6}}{4}} + c_2 (2x + 1)^{\frac{3 + \sqrt{6}}{4}}
\end{aligned}
\\
16f''(t) - 24f'(t) + 3f(t) = t\\
f(t) = At + B\\
f'(t) = A, f"(t) = 0\\
\implies -24A + 3B + 3At = t\\
\implies 3A = 1, A = \frac{1}{3} \,\,\&\,\, 3B = 24A, B = \frac{8}{3}\\
\implies f(t) = \frac{t + 8}{3}, \\
f(\ln(2x + 1)) = \frac{\ln(2x + 1) + 8}{3}\\
\implies y(x) = c_1 (2x + 1)^{\frac{3 - \sqrt{6}}{4}} + c_2 (2x + 1)^{\frac{3 + \sqrt{6}}{4}} + \frac{\ln(2x + 1) + 8}{3} 4 ( 2 x + 1 ) 2 y " − 4 ( 2 x + 1 ) y ′ + 3 y = ln ( 2 x + 1 )) y ( x ) = f ( ln ( 2 x + 1 )) y ′ ( x ) = 2 2 x + 1 f ′ ( ln ( 2 x + 1 )) y ′′ ( x ) = ( 2 x + 1 ) 2 4 f ′′ ( ln ( 2 x + 1 )) − ( 2 x + 1 ) 2 4 f ′ ( ln ( 2 x + 1 )) 16 f ′′ ( ln ( 2 x + 1 )) − 16 f ′ ( ln ( 2 x + 1 )) − 8 f ′ ( ln ( 2 x + 1 )) + 3 f ( ln ( 2 x + 1 )) = ln ( 2 x + 1 ) 16 f ′′ ( ln ( 2 x + 1 )) − 24 f ′ ( ln ( 2 x + 1 )) + 3 f ( ln ( 2 x + 1 )) = ln ( 2 x + 1 ) 16 f ′′ ( t ) − 24 f ′ ( t ) + 3 f ( t ) = t f ( t ) = c 1 e 4 3 − 6 t + c 2 e 4 3 + 6 t f ( ln ( 2 x + 1 )) = c 1 e 4 3 − 6 l n ( 2 x + 1 ) + c 2 e 4 3 + 6 l n ( 2 x + 1 ) = c 1 ( 2 x + 1 ) 4 3 − 6 + c 2 ( 2 x + 1 ) 4 3 + 6 16 f ′′ ( t ) − 24 f ′ ( t ) + 3 f ( t ) = t f ( t ) = A t + B f ′ ( t ) = A , f " ( t ) = 0 ⟹ − 24 A + 3 B + 3 A t = t ⟹ 3 A = 1 , A = 3 1 & 3 B = 24 A , B = 3 8 ⟹ f ( t ) = 3 t + 8 , f ( ln ( 2 x + 1 )) = 3 ln ( 2 x + 1 ) + 8 ⟹ y ( x ) = c 1 ( 2 x + 1 ) 4 3 − 6 + c 2 ( 2 x + 1 ) 4 3 + 6 + 3 ln ( 2 x + 1 ) + 8
#340153 on Dec 2023