Example 11. Find a particular solution of the equation

$$
y^{\prime \prime}-y=4 e^{x},
$$

satisfying the initial conditions

$$
y(0)=0, \quad y^{\prime}(0)=1
$$
Solution. We find the particular solution of equation (22)

$$
y=C_{1} e^{x}+C_{2} e^{-x}+2 x e^{x}
$$

To solve the initial value problem (22)-(23) (Cauchy problem), it is necessary to determine the values of the constants $C_{1}$ and $C_{2}$ so that the solution (24) satisfies the initial conditions (23). Using the condition $y(0)=0$, we get $C_{1}+C_{2}=0$. Differentiating (24), we find

$$
y^{\prime}=C_{1} e^{x}-C_{2} e^{-x}+2 e^{x}+2 x e^{x}
$$

from which, due to the condition $y^{\prime}(0)=1$, we have $C_{1}-C_{2}=-1$. To find $C_{1}, C_{2}$, we obtained the system

$$
\left\{\begin{array}{l}
C_{1}+C_{2}=0 \\
C_{1}-C_{2}=-1
\end{array}\right.
$$

solving which we find $C_{1}=-\frac{1}{2}, C_{2}=\frac{1}{2}$. Substituting the found values of the arbitrary constants into the general solution (24), we obtain the solution of the initial value problem (22)-(23):

$$
y=-\frac{1}{2} e^{x}+\frac{1}{2} e^{-x}+2 x e^{x} \text { or } y=2 x e^{x}-\operatorname{sh} x .
$$