Question

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

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

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
$$y = 9e^{t} + (8t + 8)e^{2}t$$