First-order linear differential equations.

First-order linear equations.

The equation $$y^{\prime }+P(x)y=Q(x)(1)$$ is called linear. To solve it, one needs to make a substitution of variables $y=u(x)v(x),$ where $u(x)$ is a solution to the homogeneous equation $u^{\prime }+P(x)u=0.$ This equation is solved by the method of separation of variables.

Next, we perform the inverse substitution. $y^{\prime }=(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }.$ Consequently,

$$u^{\prime }v+v^{\prime }u+P(x)uv=Q(x)$$

$$v(u^{\prime }+P(x)u)+v^{\prime }u=Q(x).$$

Note that $u^{\prime }+P(x)u=0.$ Consequently, we obtain an equation with separable variables.$$\begin{gathered} v^{\prime }u=Q(x) \\ {v}^{\prime }=\frac{Q(x)}{u(x)}\Rightarrow v(x)=\int \frac{Q(x)}{u(x)}dx+C. \end{gathered}$$

Some equations become linear if we swap the dependent variable and the independent variable. For example, the equation $y=(2x+y^3)y^{\prime },$ where $y$ is a function of $x,$ is nonlinear. Let's express it in differentials: $$ydx-(2x+y^3)dy=0.$$ Since $x$ and $dx$ enter linearly into this equation, the equation will be linear if we consider $x$ as the function to be found and $y$ as the independent variable. This equation can be written in the form $$\frac{dx}{dy}-\frac{2}{y}x=y^2$$ and is solved similarly to equation (1).

To solve the Bernoulli equation, that is, the equation $$y^{\prime }+a(x)y=b(y)y^n,(n\ne 1),$$ one should divide both sides by $y^n$ and make the substitution $\frac{1}{y^{n-1}}=z.$ After the substitution, a linear equation is obtained, which can be solved by the method described above.

Riccati's equation.

The Riccati equation, i.e., the equation $$y^{\prime }+a(x)y+b(x)y^2=c(x),$$ is not solvable in quadratures in the general case. However, if one particular solution $y_1(x)$ is known, then by substituting $y=y_1(x)+z,$ the Riccati equation is reduced to a Bernoulli equation and thus can be solved in quadratures.

Sometimes it is convenient to choose a particular solution based on the form of the free term of the equation (the term that does not contain $y$). For example, for the equation $y^{\prime }+y^2=x^2-2x$, in the left side, there will be terms similar to the terms in the right side if we take $y=ax+b.$ By substituting into the equation and equating coefficients at similar terms, we find $a$ and $b$ (if a particular solution of the specified form exists, which is not always the case). Another example: for the equation $y^{\prime }+2y^2=\frac{6}{x^2},$ the same reasoning prompts us to look for a particular solution in the form $y=\frac{a}{x}.$ By substituting $y=\frac{a}{x}$ into the equation, we find the constant $a.$

Tags: Riccati's equation, differential equations, bernoulli equation, first-order linear equations