Answer to Question #281848 in Differential Equations for Rasna

Question

Use the method of variation of parameters to solve the differential equation:

y"-2y'+y=xe^xtan-1x

Answer 1

characteristic equation:

$k^2-2k+1=0$

$k_{1,2}=1$

complementary solution:

$y_c=c_1e^x+c_2xe^x$

particular solution:

$y_p=-y_1\int\frac{y_2g(x)}{W}dx+y_2\int\frac{y_1g(x)}{W}dx$

where $y_1=e^x,y_2=xe^x$

$W=y_1y'_2-y'_1y_2=e^x(e^x+xe^x)-xe^{2x}=e^{2x}$

$g(x)=xe^xtan^{-1}x$

$\int\frac{y_2g(x)}{W}dx=\int\frac{x^2e^{2x}tan^{-1}x}{e^{2x}}dx=\frac{ln(x^2+1)+2x^3tan^{-1}x-x^2}{6}$

$\int\frac{y_1g(x)}{W}dx=\int\frac{xe^{2x}tan^{-1}x}{e^{2x}}dx=\frac{(x^2+1)tan^{-1}x-x}{2}$

$y=y_c+y_p=c_1e^x+c_2xe^x-e^x\frac{ln(x^2+1)+2x^3tan^{-1}x-x^2}{6}+xe^x\frac{(x^2+1)tan^{-1}x-x}{2}$