Respuesta :
Answer:
Therefore the complete primitive is
[tex]y=c_1 e^{2y}+c_2e^{3t}+e^{t}[/tex]
Therefore the general solution is
[tex]y=c_1e^{2t}+c_2e^{3t}+e^t[/tex]
Step-by-step explanation:
Given Differential equation is
[tex]y''-5y'+6y=2e^t[/tex]
Method of variation of parameters:
Let [tex]y=e^{mt}[/tex] be a trial solution.
[tex]y'= me^{mt}[/tex]
and [tex]y''= m^2e^{mt}[/tex]
Then the auxiliary equation is
[tex]m^2e^{mt}-5me^{mt}+6e^{mt}=0[/tex]
[tex]\Rightarrow m^2-5m+6=0[/tex]
[tex]\Rightarrow m^2 -3m -2m +6=0[/tex]
[tex]\Rightarrow m(m -3) -2(m -3)=0[/tex]
[tex]\Rightarrow (m-3)(m-2)=0[/tex]
[tex]\Rightarrow m=2,3[/tex]
∴The complementary function is [tex]C_1e^{2t}+C_2e^{3t}[/tex]
To find P.I
First we show that [tex]e^{2t}[/tex] and [tex]e^{3t}[/tex] are linearly independent solution.
Let [tex]y_1=e^{2t}[/tex] and [tex]y_2= e^{3t}[/tex]
The Wronskian of [tex]y_1[/tex] and [tex]y_2[/tex] is [tex]\left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|[/tex]
[tex]=\left|\begin{array}{cc}e^{2t}&e^{3t}\\2e^{2t}&3e^{3t}\end{array}\right|[/tex]
[tex]=e^{2t}.3e^{3t}-e^{2t}.2e^{3t}[/tex]
[tex]=e^{5t}[/tex] ≠ 0
∴[tex]y_1[/tex] and [tex]y_2[/tex] are linearly independent.
Let the particular solution is
[tex]y_p=v_1(t)e^{2t}+v_2(t)e^{3t}[/tex]
Then,
[tex]Dy_p= 2v_1(t)e^{2t}+v'_1(t)e^{2t}+3v_2(t)e^{3t}+v'_2(t)e^{3t}[/tex]
Choose [tex]v_1(t)[/tex] and [tex]v_2(t)[/tex] such that
[tex]v'_1(t)e^{2t}+v'_2(t)e^{3t}=0[/tex] .......(1)
So that
[tex]Dy_p= 2v_1(t)e^{2t}+3v_2(t)e^{3t}[/tex]
[tex]D^2y_p= 4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}[/tex]
Now
[tex]4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}-5[2v_1(t)e^{2t}+3v_2(t)e^{3t}] +6[v_1e^{2t}+v_2e^{3t}]=2e^t[/tex]
[tex]\Rightarrow 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}=2e^t[/tex] .......(2)
Solving (1) and (2) we get
[tex]v'_2=2 e^{-2t}[/tex] and [tex]v'_1(t)=-2e^{-t}[/tex]
Hence
[tex]v_1(t)=\int (-2e^{-t}) dt=2e^{-t}[/tex]
and [tex]v_2=\int 2e^{-2t}dt =-e^{-2t}[/tex]
Therefore [tex]y_p=(2e^{-t}) e^{2t}-e^{-2t}.e^{3t}[/tex]
[tex]=2e^t-e^t[/tex]
[tex]=e^t[/tex]
Therefore the complete primitive is
[tex]y=c_1 e^{2y}+c_2e^{3t}+ e^{t}[/tex]
Undermined coefficients:
∴The complementary function is [tex]C_1e^{2t}+C_2e^{3t}[/tex]
The particular solution is [tex]y_p=Ae^t[/tex]
Then,
[tex]Dy_p= Ae^t[/tex] and [tex]D^2y_p=Ae^t[/tex]
[tex]\therefore Ae^t-5Ae^t+6Ae^t=2e^t[/tex]
[tex]\Rightarrow 2Ae^t=2e^t[/tex]
[tex]\Rightarrow A=1[/tex]
[tex]\therefore y_p=e^t[/tex]
Therefore the general solution is
[tex]y=c_1e^{2t}+c_2e^{3t}+e^t[/tex]
