Example. Find the general solution of the differential equation

$$
\left(1+x^{2}\right) y^{\prime \prime}+2 x y^{\prime}=12 x^{3}
$$
Solution.

1. Since the differential equation does not contain $y$, by setting $y^{\prime}=p(x)$, we have $y^{\prime \prime}=p^{\prime}(x)$. We obtain a first-order differential equation

$$
\left(1+x^{2}\right) p^{\prime}+2 x p=12 x^{3}
$$

## 2. The equation

$$
p^{\prime}+\frac{2 x}{1+x^{2}} p=\frac{12 x^{3}}{1+x^{2}}
$$

is linear with respect to $p$ and $p^{\prime}$. Solving it, for example, by the method of variation of arbitrary constants, we find

$$
p=\frac{3 x^{4}+C}{1+x^{2}}
$$

3. Since $p=y^{\prime}(x)$, we have

$$
y^{\prime}=\frac{3 x^{4}+C}{1+x^{2}}
$$

Integrating, we obtain the general solution $y=(C+3) \operatorname{arctg} x+x^{3}-3 x+C_{2}$. Answer. $y=C_{1} \operatorname{arctg} x+x^{3}-3 x+C_{2}$.

Conditions of the Problems. Find the general solutions of the differential equations.

1. $y^{\prime \prime}=1-{y^{\prime}}^{2}$.
2. $x y^{\prime \prime}+y^{\prime}=0$.
3. $\left(1+x^{2}\right) y^{\prime \prime}+y^{\prime 2}+1=0$.
4. $x^{2} y^{\prime \prime}+x y^{\prime}=1$.
5. $x y^{\prime \prime \prime}+y^{\prime \prime}=1+x$.
6. $y^{\prime \prime \prime 2}+y^{\prime \prime \prime}=1$.
7. $y^{\prime}\left(1+y^{\prime 2}\right)=y^{\prime \prime}$.
8. $y^{\prime \prime}=-x / y^{\prime}$.
9. $x y^{\prime \prime}+y^{\prime}+x=0$.
10. $y^{\prime \prime \prime 2}=4 y^{\prime \prime}$.

Answers.

1. $y=\ln \left|e^{2 x}+C_{1}\right|-x+C_{2}$.
2. $y=C_{1}+C_{2} \ln |x|$.
3. $y=\left(1+C_{1}^{2}\right) \ln \left|x+C_{1}\right|-C_{1} x+C_{2}$.
4. $y=\frac{1}{2}(\ln |x|)^{2}+C_{1} \ln |x|+C_{2}$.
5. $y=\frac{x^{3}}{12}+\frac{x^{2}}{2}+C_{1} x \ln |x|+C_{2} x+C_{3}$.
6. $y=\sin \left(C_{1}+x\right)+C_{2} x+C_{3}$.
7. $x-C_{1}=\ln \left|\sin \left(y-C_{2}\right)\right|$.
8. $y=C_{1} x^{2}\left(\ln x-\frac{3}{2}\right)+C_{2} x+C_{3}$.
9. $y=C_{1} \ln |x|-\frac{x^{2}}{4}+C_{2}$.
10. $y=\frac{1}{2}\left(C_{1}+x\right)^{4}+C_{2} x+C_{3}$.

## 11.8. Equations of the form $F\left(y, y^{\prime}, y^{\prime \prime}\right)=0$

Problem Statement. Find the solution to the Cauchy problem for the differential equation

$$
F\left(y, y^{\prime}, y^{\prime \prime}\right)=0
$$

with initial conditions

$$
y\left(x_{0}\right)=y_{0}, \quad y^{\prime}\left(x_{0}\right)=y_{0}^{\prime} .
$$

PLAN OF SOLUTION.

1. Since the differential equation does not explicitly contain the independent variable $x$, we set

$$
y^{\prime}=p(y)
$$

where $p(y)$ is a new unknown function. Then, by the chain rule for the derivative of a composite function, we have

$$
y^{\prime \prime}=\frac{d}{d x} y^{\prime}=\frac{d}{d x} p(y)=\frac{d p}{d y} \cdot \frac{d y}{d x}=p \frac{d p}{d y}
$$

We obtain a first-order equation in terms of $p(y)$

$$
F\left(y, p, p \frac{d p}{d y}\right)=0
$$

2. By identifying the type of this equation and applying the corresponding method of solution, we find $p=f(y, C)$, where $C$ is an arbitrary constant.
3. Using the initial conditions (both), we find $C=C_{1}$.
4. Substituting $C_{1}$, we obtain a differential equation with separable variables

$$
\frac{d y}{d x}=f\left(y, C_{1}\right)
$$

Separating the variables in the region where $f\left(y, C_{1}\right) \neq 0$, we get

$$
\frac{d y}{f\left(y, C_{1}\right)}=d x
$$

and, integrating, we find $x=\varphi\left(y, C_{1}, C_{2}\right)$.

We check whether the solution $f\left(y, C_{1}\right)=0$ is a particular solution of the original equation that satisfies the initial conditions.

5. Using the initial conditions to find the second constant $C_{2}$ (the value of $C_{1}$ has already been found in step 3) and we obtain the solution to the Cauchy problem.

The answer is written in the form $y=y(x)$ or $x=x(y)$.