Respuesta :
Answer:
[tex]y(t)\ =\ C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]
Step-by-step explanation:
As given in question, we have to find the solution of differential equation
[tex]y"+y'-2y=1[/tex]
by using the variation in parameter method.
From the above equation, the characteristics equation can be given by
[tex]D^2+D-2\ =\ 0[/tex]
[tex]=>D=\ \dfrac{-1+\sqrt{1^2+4\times 2\times 1}}{2\times 1}\ or\ \dfrac{-1-\sqrt{1^2+4\times 2\times 1}}{2\times 1}[/tex]
[tex]=>\ D=\ -2\ or\ 1[/tex]
Since, the roots of characteristics equation are real and distinct, so the complementary function of the differential equation can be by
[tex]y_c(t)\ =\ C_1e^{-2t}+C_2e^t[/tex]
Let's assume that
[tex]y_1(t)=e^{-2t}[/tex] [tex]y_2(t)=e^t[/tex]
[tex]=>\ y'_1(t)=-2e^{-2t}[/tex] [tex]y'_2(t)=e^t[/tex]
and g(t)=1
Now, the Wronskian can be given by
[tex]W=y_1(t).y'_2(t)-y'_1(t).y_2(t)[/tex]
[tex]=e^{-2t}.e^t-e^t(-e^{-2t})[/tex]
[tex]=e^{-t}+2e^{-t}[/tex]
[tex]=3e^{-t}[/tex]
Now, the particular solution can be given by
[tex]y_p(t)\ =\ -y_1(t)\int{\dfrac{y_2(t).g(t)}{W}dt}+y_2(t)\int{\dfrac{y_1(t).g(t)}{W}dt}[/tex]
[tex]=\ -e^{-2t}\int{\dfrac{e^t.1}{3.e^{-t}}dt}+e^{t}\int{\dfrac{e^{-2t}.1}{3.e^{-t}}dt}[/tex]
[tex]=\ -e^{-2t}\int{\dfrac{1}{3}dt}+\dfrac{e^t}{3}\int{e^{-t}dt}[/tex]
[tex]=\dfrac{-e^{-2t}}{3}.t-\dfrac{1}{3}[/tex]
[tex]=-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]
Now, the complete solution of the given differential equation can be given by
[tex]y(t)\ =\ y_c(t)+y_p(t)[/tex]
[tex]=C_1e^{-2t}+C_2e^t-t\dfrac{e^{-2t}}{3}-\dfrac{1}{3}[/tex]
