120. Find the solution of the equation $y^{\prime \prime}=2$, satisfying the following initial conditions: $y=1$ and $y^{\prime}=2$ when $x=0$.
Solution. Integrating the given equation with respect to $x$ twice in succession, we first obtain $y^{\prime}=2 x+C_{1}$, and then $y=x^{2}+C_{1} x+C_{2}$. Using the initial conditions now, we have

$$
\left\{\begin{array}{l}
y^{\prime}=2 x+C_{1} \\
y=x^{2}+C_{1} x+C_{2};
\end{array} \quad \left\{\begin{array}{l}
2=2 \cdot 0+C_{1} \\
1=0+C_{1} \cdot 0+C_{2};
\end{array} \left\{\begin{array}{l}
2=C_{1} \\
1=C_{2}
\end{array}\right.\right.\right.
$$

i.e., $C_{1}=2, C_{2}=1$. Therefore, the desired solution is $y=x^{2}+2 x+1$ or $y=(x+1)^{2}$.

Thus, to solve a differential equation of the form $y^{\prime \prime}=f(x)$, the following algorithm is used:

$1^{0}$. Integrate both sides of the equation to find $\frac{d y}{d x}$.

$2^{0}$. By integrating $\frac{d y}{d x}$, find the general solution containing two arbitrary constants.

$3^{0}$. If a particular solution (solution to the Cauchy problem) is required, determine $C_{1}$ and $C_{2}$ from the initial conditions and substitute them into the general solution.