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

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

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

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

Matriz Wroskiana

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