# Solve differential equationy '(t) = -3y + 9, y(0) = 4

Solve differential equation $$y '(t) = -3y + 9$$, $$y(0) = 4$$

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casincal

$$dy/dx= -3y+9$$
$$dy/dx+3y= 9$$
$$I.F.= e^{\int\ 3dt}= e^{3t}$$
$$D_t(e^{3t}y)= 9e^{3t}$$
$$e^{3t}y=0 \int e^{3t}dt$$
$$e^{3t}y=27e^{3t}+C$$
$$y(t)= 27+Ce^{-3t}$$
now as $$y(0)=4$$
$$y(0)= 27+Ce^0$$
$$C= 4-27$$
$$C= -23$$
$$y(t)= 27-23e^{-3t}$$