Question

Find the sokaton 02 the ger Initial value provsem. y′−y=8t2,y(0)=1

First order differential equations
ANSWERED
asked 2020-11-10
Find the sokaton 02 the ger Initial value provsem. \(y′−y=8t2,y(0)=1\)

Answers (1)

2020-11-11

We first solve the homogeneous equation:
\(y` - y = 0\)
Its characteristic equation is
\(r - 1 = 0 \Rightarrow r = 1\)
Therefore, the solution of the homogeneous equation is
\(y_{H} = Ce^{t}\)
Now we need to find a particular solution. We will try with
\(y_{p}=(At+B)e^{2}t \Rightarrow y'_{p}, =2(At+B)e^{2}t+Ae^{2}t=(2At+(A+2B))e^{2}t\)
Thus,
\(y`_{p} - y_{p} = 8te^{2}t \Rightarrow (2At + (A + 2B))e^{2}t - (At + B)e^{2}t = 8te^{2}t \Rightarrow (At + (A + B))e^{2}t = 8te^{2}t\)
From this we get
\(At + (A + B) = 8t,\)
and the system
\(A = 8\)
\(A + B = 0\)
Therefore,
\(A = 8 B = -A = -8\)
and the particular solution is
\(y_{p} = (8t - 8)e^{2}t\)
This means that the total solution is
\(y = y_{H} + y_{p} = Ce^{t} + (8t - 8)e^{2}t\)
Now we need to find C. We use the fact that y(0) = 1, so
\(1 = y(0) = Ce^{0} + (8 \cdot 0 - 8)e^{2 \cdot 0} = C - 8\)
Therefore, \(C = 9\)
and the solution of the initial equation is
\(y = 9e^{t} + (8t + 8)e^{2}t\)

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