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-∫0x(x-s)f(s)ds where c1 and c2 are arbitrary constants.
y"=−f(x)y"=0y′=C2y=C2+C1xWronskian(1,x)=1C2=−∫x×−f(x)1dx=∫xf(x) dx+c2C1=∫1×−f(x)1dx=−∫f(x) dx+c1y=−x(∫f(x) dx+c1)+∫xf(x) dx+c2=c1x+c2−x(∫0xf(s) ds)+∫0xsf(s) ds=c1x+c2−∫0xxf(s) ds+∫0xsf(s) ds=c1x+c2−∫0x(x−s)f(s) ds\displaystyle y" = -f(x)\\ y" = 0\\ y' = C_2\\ y = C_2 + C_1x\\ \textsf{Wronskian}(1, x) = 1\\ C_2 = -\int \frac{x \times -f(x)}{1} \mathrm{d}x = \int xf(x)\,\,\mathrm{d}x + c_2\\ C_1 = \int \frac{1 \times -f(x)}{1} \mathrm{d}x = -\int f(x)\,\,\mathrm{d}x + c_1\\ \begin{aligned} y &= -x\left(\int f(x)\,\,\mathrm{d}x + c_1\right) + \int xf(x)\,\,\mathrm{d}x + c_2 \\&= c_1x + c_2 - x\left(\int_0^x f(s)\,\,\mathrm{d}s\right) + \int_0^x sf(s)\,\,\mathrm{d}s \\&= c_1x + c_2 - \int_0^x xf(s)\,\,\mathrm{d}s + \int_0^x sf(s)\,\,\mathrm{d}s \\&= c_1x + c_2 - \int_0^x (x - s)f(s)\,\,\mathrm{d}s \end{aligned}y"=−f(x)y"=0y′=C2y=C2+C1xWronskian(1,x)=1C2=−∫1x×−f(x)dx=∫xf(x)dx+c2C1=∫11×−f(x)dx=−∫f(x)dx+c1y=−x(∫f(x)dx+c1)+∫xf(x)dx+c2=c1x+c2−x(∫0xf(s)ds)+∫0xsf(s)ds=c1x+c2−∫0xxf(s)ds+∫0xsf(s)ds=c1x+c2−∫0x(x−s)f(s)ds
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