Answer to Question #224553 in Differential Equations for smi

Question

Show by the method of variation of parameters that the general solution of the differential equation -y^''=f(x) can be written in the form y = φ ( x) = c₁+c₂x-∫₀^x(x-s)f(s)ds where c₁ and c₂ are arbitrary constants.

Answer 1

"\\displaystyle\ny" = -f(x)\\\\\ny" = 0\\\\\ny' = C_2\\\\\ny = C_2 + C_1x\\\\\n\n\\textsf{Wronskian}(1, x) = 1\\\\\n\nC_2 = -\\int \\frac{x \\times -f(x)}{1} \\mathrm{d}x = \\int xf(x)\\,\\,\\mathrm{d}x + c_2\\\\\n \nC_1 = \\int \\frac{1 \\times -f(x)}{1} \\mathrm{d}x = -\\int f(x)\\,\\,\\mathrm{d}x + c_1\\\\\n\n\\begin{aligned}\ny &= -x\\left(\\int f(x)\\,\\,\\mathrm{d}x + c_1\\right) + \\int xf(x)\\,\\,\\mathrm{d}x + c_2\n\\\\&= c_1x + c_2 - x\\left(\\int_0^x f(s)\\,\\,\\mathrm{d}s\\right) + \\int_0^x sf(s)\\,\\,\\mathrm{d}s\n\\\\&= c_1x + c_2 - \\int_0^x xf(s)\\,\\,\\mathrm{d}s + \\int_0^x sf(s)\\,\\,\\mathrm{d}s\n\\\\&= c_1x + c_2 - \\int_0^x (x - s)f(s)\\,\\,\\mathrm{d}s\n\\end{aligned}"