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

Find the sokaton 02 the ger Initial value provsem. $y\prime -y=8t2,y\left(0\right)=1$
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We first solve the homogeneous equation:
$y‘-y=0$
Its characteristic equation is
$r-1=0⇒r=1$
Therefore, the solution of the homogeneous equation is
${y}_{H}=C{e}^{t}$
Now we need to find a particular solution. We will try with
${y}_{p}=\left(At+B\right){e}^{2}t⇒{y}_{p}^{\prime },=2\left(At+B\right){e}^{2}t+A{e}^{2}t=\left(2At+\left(A+2B\right)\right){e}^{2}t$
Thus,
$y{‘}_{p}-{y}_{p}=8t{e}^{2}t⇒\left(2At+\left(A+2B\right)\right){e}^{2}t-\left(At+B\right){e}^{2}t=8t{e}^{2}t⇒\left(At+\left(A+B\right)\right){e}^{2}t=8t{e}^{2}t$
From this we get
$At+\left(A+B\right)=8t,$
and the system
$A=8$
$A+B=0$
Therefore,
$A=8B=-A=-8$
and the particular solution is
${y}_{p}=\left(8t-8\right){e}^{2}t$
This means that the total solution is
$y={y}_{H}+{y}_{p}=C{e}^{t}+\left(8t-8\right){e}^{2}t$
Now we need to find C. We use the fact that y(0) = 1, so
$1=y\left(0\right)=C{e}^{0}+\left(8\cdot 0-8\right){e}^{2\cdot 0}=C-8$
Therefore, $C=9$
and the solution of the initial equation is
$y=9{e}^{t}+\left(8t+8\right){e}^{2}t$