Question

y'' -3 y'+2у=sinx-7cosx y(0)=2 y'(0)=7

y'' -10 y'+25у=е(Степени) 5х y(0)=1 y'(0)=0

Answer 1

$y''-3y'+2y=sin x-7cos x$

resolvamos la EDO homogénea $y''-3y'+2y=0$

           $r^2-3r+2=0to (r-1)(r-2)=0to r=1vee r=2\ \ boxed{y_c=C_1e^x+C_2e^{2x}}$

Matriz Wroskiana

        $left[begin{array}{ccc} e^x&e^{2x}\e^x&2e^{2x} end{array}right] left[begin{array}{ccc} u_1'\u_2' end{array}right]=left[begin{array}{ccc} 0\sin x-7cos x end{array}right] \ displaystyle\ \ u_1'=e^{-x}(7cos x-sin x)to boxed{u_1=e^{-x}(4sin x-3cos x)}\ \ u_2'=e^{-2x}(sin x-7cos x)to boxed{u_2=e^{-2x}left(dfrac{13cos x}{5}-dfrac{9sin x}{5}right)}\ \ \ y=C_1e^x+C_2e^{2x}+u_1e^x+u_2e^{2x}\ \ boxed{y=C_1e^x+C_2e^{2x}- dfrac{2cos x}{5}+dfrac{11sin x}{5}}$