4.92. Find the solution to the Cauchy problem for the equation of the form
$y^{(n)}(x)=F(x): y^{\prime \prime \prime}=24 x, y(0)=3, y^{\prime}(0)=2, y^{\prime \prime}(0)=4$.
Solution. After the first integration, we get $y^{\prime \prime}(x)=\int 24 x d x+C_{1}=24 \cdot \frac{x^{2}}{2}+C_{1}$. After the second integration, we find $y^{\prime}(x)=\int\left(12 x^{2}+C_{1}\right) d x=12 \cdot \frac{x^{3}}{3}+C_{1} x+C_{2}$. Finally, after the third integration, we write the general solution:

$$
y(x)=\int\left(4 x^{3}+C_{1} x+C_{2}\right) d x=4 \cdot \frac{x^{4}}{4}+C_{1} \frac{x^{2}}{2}+C_{2} x+C_{3}
$$

Using the obtained expressions for $y^{\prime \prime}(x), y^{\prime}(x)$, and $y(x)$, and with the initial conditions, we find: $C_{1}=4, C_{2}=2$, $C_{3}=3$. The solution to the Cauchy problem is:

$$
y(x)=x^{4}+2 x^{2}+2 x+3
$$