Question Number 125651 by Algoritm last updated on 12/Dec/20
Answered by Olaf last updated on 12/Dec/20
![{ ((y_1 ′ = y_1 +4y_2 (1))),((y_2 ′ = −y_1 +y_2 +e^(3x) (2))) :} (2) : y_1 = y_2 −y_2 ′+e^(3x) (3) ⇒ y_1 ′ = y_2 ′−y_2 ′′+3e^(3x) (1) y_2 ′−y_2 ′′+3e^(3x) = y_2 −y_2 ′+e^(3x) +4y_2 y_2 ′′−2y_2 ′+5y_2 = 2e^(3x) (E_H ) : y_(2H) ′′−2y_(2H) +5y_(2H) = 0 r^2 −2r+5 = 0 r = ((2±4i)/2) = 1±2i y_(2H) = e^x (Acos2x+Bsin2x) (E_0 ) : y_(2,0) ′′−2y_(2,0) +5y_(2,0) = e^(3x) y_(2,0) = αe^(3x) ⇒ 9α−2×3α+5α = 2 α = (1/4), y_(2,0) = (e^(3x) /4) y_2 = y_(2H) +y_(2,0) = e^x (Acos2x+Bsin2x)+(e^(3x) /4) ⇒ y_2 ′ = e^x [(A+2B)cos2x+(B−2A)sin2x]+((3e^(3x) )/4) (3) : y_1 = −2e^x (Bcos2x−Asin2x)−(e^(3x) /2)](https://www.tinkutara.com/question/Q125653.png)
$$\begin{cases}{{y}_{\mathrm{1}} '\:=\:{y}_{\mathrm{1}} +\mathrm{4}{y}_{\mathrm{2}} \:\left(\mathrm{1}\right)}\\{{y}_{\mathrm{2}} '\:=\:−{y}_{\mathrm{1}} +{y}_{\mathrm{2}} +{e}^{\mathrm{3}{x}} \:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{2}\right)\::\:{y}_{\mathrm{1}} \:=\:{y}_{\mathrm{2}} −{y}_{\mathrm{2}} '+{e}^{\mathrm{3}{x}} \:\left(\mathrm{3}\right) \\ $$$$\Rightarrow\:{y}_{\mathrm{1}} '\:=\:{y}_{\mathrm{2}} '−{y}_{\mathrm{2}} ''+\mathrm{3}{e}^{\mathrm{3}{x}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\:{y}_{\mathrm{2}} '−{y}_{\mathrm{2}} ''+\mathrm{3}{e}^{\mathrm{3}{x}} \:=\:{y}_{\mathrm{2}} −{y}_{\mathrm{2}} '+{e}^{\mathrm{3}{x}} +\mathrm{4}{y}_{\mathrm{2}} \\ $$$${y}_{\mathrm{2}} ''−\mathrm{2}{y}_{\mathrm{2}} '+\mathrm{5}{y}_{\mathrm{2}} \:=\:\mathrm{2}{e}^{\mathrm{3}{x}} \\ $$$$\left(\mathrm{E}_{\mathrm{H}} \right)\::\:{y}_{\mathrm{2H}} ''−\mathrm{2}{y}_{\mathrm{2H}} +\mathrm{5}{y}_{\mathrm{2H}} \:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} −\mathrm{2}{r}+\mathrm{5}\:=\:\mathrm{0} \\ $$$${r}\:=\:\frac{\mathrm{2}\pm\mathrm{4}{i}}{\mathrm{2}}\:=\:\mathrm{1}\pm\mathrm{2}{i} \\ $$$${y}_{\mathrm{2H}} \:=\:{e}^{{x}} \left(\mathrm{Acos2}{x}+\mathrm{Bsin2}{x}\right) \\ $$$$\left(\mathrm{E}_{\mathrm{0}} \right)\::\:{y}_{\mathrm{2},\mathrm{0}} ''−\mathrm{2}{y}_{\mathrm{2},\mathrm{0}} +\mathrm{5}{y}_{\mathrm{2},\mathrm{0}} \:=\:{e}^{\mathrm{3}{x}} \\ $$$${y}_{\mathrm{2},\mathrm{0}} \:=\:\alpha{e}^{\mathrm{3}{x}} \\ $$$$\Rightarrow\:\mathrm{9}\alpha−\mathrm{2}×\mathrm{3}\alpha+\mathrm{5}\alpha\:=\:\mathrm{2} \\ $$$$\alpha\:=\:\frac{\mathrm{1}}{\mathrm{4}},\:{y}_{\mathrm{2},\mathrm{0}} \:=\:\frac{{e}^{\mathrm{3}{x}} }{\mathrm{4}} \\ $$$${y}_{\mathrm{2}} \:=\:{y}_{\mathrm{2H}} +{y}_{\mathrm{2},\mathrm{0}} \:=\:{e}^{{x}} \left(\mathrm{Acos2}{x}+\mathrm{Bsin2}{x}\right)+\frac{{e}^{\mathrm{3}{x}} }{\mathrm{4}} \\ $$$$\Rightarrow\:{y}_{\mathrm{2}} '\:=\:{e}^{{x}} \left[\left(\mathrm{A}+\mathrm{2B}\right)\mathrm{cos2}{x}+\left(\mathrm{B}−\mathrm{2A}\right)\mathrm{sin2}{x}\right]+\frac{\mathrm{3}{e}^{\mathrm{3}{x}} }{\mathrm{4}} \\ $$$$\left(\mathrm{3}\right)\::\:{y}_{\mathrm{1}} \:=\:−\mathrm{2}{e}^{{x}} \left(\mathrm{Bcos2}{x}−\mathrm{Asin2}{x}\right)−\frac{{e}^{\mathrm{3}{x}} }{\mathrm{2}} \\ $$