189. Find the integral curve of the differential equation $y^{\prime \prime}+2 y^{\prime}+2 y=0$, passing through the point $(0 ; 1)$ and tangent to the line $y=x+1$ at this point.
Solution. The roots of the characteristic equation $k^{2}+2 k+2=$ $=0$ are complex conjugates: $-1 \pm i$. The equation of the family of integral curves of the given differential equation is written in the form

$$
y=e^{-x}\left(C_{1} \cos x+C_{2} \sin x\right)
$$

To find the equation of the desired integral curve, substitute into the equalities

$$
y=e^{-x}\left(C_{1} \cos x+C_{2} \sin x\right)
$$

and

$$
y^{\prime}=e^{-x}\left(\left(C_{2}-C_{1}\right) \cos x-\left(C_{2}+C_{1}\right) \sin x\right)
$$

the values of the ordinate $y=1$ and the slope of the tangent $y^{\prime}=k=1$ at the point $x=0$. We obtain the system of equations $1=C_{1}, 1=$ $=C_{2}-C_{1}$, from which we determine $C_{1}=1, C_{2}=2$. Substituting these values into the equation of the family, we obtain the desired equation

$$
y=e^{-x}(\cos x+2 \sin x)
$$