121. Solve the Cauchy problem for the equation $y^{\prime \prime}=1+x+x^{2}+$ $+x^{3}$, if $y=1$ and $y^{\prime}=1$ when $x=0$.
Solution. $1^{0}$. $\frac{d y}{d x}=\int\left(1+x+x^{2}+x^{3}\right) d x=x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+C_{1}$.

$2^{0} \cdot y=\int\left(x+\frac{x^{2}}{2}+\frac{x^{3}}{3}+\frac{x^{4}}{4}+C_{1}\right) d x=\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+\frac{x^{5}}{20}+C_{1} x+C_{2}$.
$3^{0}$. Substituting the initial conditions $y=1$ and $y^{\prime}=1$ at $x=0$, we get $C_{1}=1, C_{2}=1$.

Therefore, $y=\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+\frac{x^{5}}{20}+x+1$.

122-127. Solve the Cauchy problem for the equations: