# Answer in Differential Equations

“displaystyle x^2y”- 2xy’ + 2y=x^2+sin(ln(5x))\nnx = e^t\nne^{2t} y” – 2e^{t}y’ + 2y = e^{2t} + sin(5t) \nnny = f(ln(x)) \nnnfrac{mathrm{d}y}{mathrm{d}t} = frac{f'(ln(x))}{x}\nnnfrac{mathrm{d}^2y}{mathrm{d}t^2} = frac{-f'(ln(x))}{x^2} + frac{f”(ln(x))}{x^2} = frac{f”(ln(x)) – f'(ln(x))}{x^2}\nnnne^{2t}left(frac{f”(ln(x)) – f'(ln(x))}{x^2}right) – 2e^{t}frac{f'(ln(x))}{x} + 2f(ln(x)) = e^{2t} + sin(5t) \nnne^{2t}left(frac{f”(ln(x)) – f'(ln(x))}{e^{2t}}right) – 2e^{t}frac{f'(ln(x))}{e^t} + 2f(ln(x)) = e^{2t} + sin(5t)\nnnf”(ln(x)) – f'(ln(x)) – 2f'(ln(x)) + 2f(ln(x)) = e^{2t} + sin(5t) \nnnf”(ln(x)) – 3f'(ln(x)) + 2f(ln(x)) = e^{2t} + sin(5t) \nnnnf”(t) – 3f'(t) + 2f(t) = e^{2t} + sin(5t) \nnf(t) = f_p + f_c\nnnf_c , textsf{is the complementary factor while}\f_p , textsf{is the particular integral}\nntextsf{The auxiliary equation is}\nnm^2 – 3m + 2 = 0\nn(m – 1)(m – 2) = 0\nnm = 1, 2\nnnf”(t) – 3f'(t) + 2f(t) = e^{2t} + sin(5t)\nnntextsf{Solving by the method of variation of parameters}\nnW(t) , textsf{is the Wronskian of the solution to the DE}\nnW(t) = e^{-int -3 , mathrm{d}t} = e^{3t}\nnnV_1(t) = -int e^t frac{(e^{2t} + sin(5t))}{e^{3t}} ,mathrm{d}t = -int (1 + e^{-2t}sin(5t))) ,mathrm{d}t\nnnint e^{-2t}sin(5t) , mathrm{d}t = -frac{2e^{-2t}sin(5t)}{29} -frac{5e^{-2t}cos(5t)}{29}\nnnV_1(t) = -t + frac{2e^{-2t}sin(5t)}{29} + frac{5e^{-2t}cos(5t)}{29}\nnbegin{aligned}nV_2(t) &= int e^{2t} frac{(e^{2t} + sin(5t))}{e^{3t}} ,mathrm{d}t \&= int e^t + e^{-t}sin(5t) ,mathrm{d}t \&= e^t – frac{5}{26}e^{-t} cos(5t) – frac{1}{26} e^{-t}sin(5t)\nend{aligned}\nnbegin{aligned}nf_p &= left(-t + frac{2e^{-2t}sin(5t)}{29} + frac{5e^{-2t}cos(5t)}{29}right)e^{2t} \&+ left(e^t – frac{5}{26}e^{-t} cos(5t) – frac{1}{26} e^{-t}sin(5t)right)e^t \&= -te^{2t}+ frac{2sin(5t)}{29} + frac{5cos(5t)}{29} + e^{2t} – frac{5}{26} cos(5t) – frac{1}{26}sin(5t) \&= e^{2t}(1 – t) – frac{15}{784}cos(5t) + frac{23}{754}sin(5t)\nend{aligned}\nnnf”(t) – 3f'(t) + 2f(t) = e^{2t} + sin(5t)\nnntherefore f(t) = Ae^t + Be^{2t} + e^{2t}(1 – t) – frac{15}{784}cos(5t) + frac{23}{754}sin(5t)\nnntherefore y = f(x) = Ax + Bx^2 + x^2(1 – ln(x)) – frac{15}{784}cos(5ln(x)) + frac{23}{754}sin(5ln(x))\nntextsf{Is the solution to the ODE}”

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