Differential equations with separable variables and those reducible to them.

Equations with separable variables.

Equations with separable variables can be written in the form$$y^{\prime }=f(x)g(y),(1)$$ and also in the form $$M(x)N(y)dx+P(x)Q(y)dy=0.(2)$$ o solve this equation, it needs to be transformed in such a way that one part of the equation involves only $x$, and the other only involves $y$, and then integrate both parts.

When dividing both sides of the equation by an expression containing the unknowns $x$ and $y$, solutions that make this expression zero may be lost.

Example. Solve the equation $$x^2{y}^2{y}^{\prime }+1=y.$$

Solution.

Let's transform the given equation into the form (2).

$$x^2{y}^2\frac{dx}{dy}=y-1;x^2{y}^{2}dy=(y-1)dx.$$

We divide both sides of the equation by $x^2(y-1):$

$$\frac{y^2}{y-1}dy=\frac{dx}{x^2}.$$

The variables are separated. We integrate both sides of the equation:

$$\int \frac{y^2}{y-1}dy=\int \frac{dx}{x^2}.$$

$$\int \frac{y^2}{y-1}dy=\int \frac{y^2-1+1}{y-1}dy=\int (y+1+\frac{1}{y-1})dy=\frac{y^2}{2}+y+\ln |y-1|+C.$$

$$\int \frac{dx}{x^2}=-\frac{1}{x}+C.$$

Thus,

$$\frac{y^2}{2}+y+\ln |y-1|=-\frac{1}{x}+C.$$

When dividing by $x^2(y-1)$, solutions $x=0$ and $y-1=0,$ i.e., $y=1,$ could have been lost. If we substitute these values into the original equation, it becomes evident that $y=1$ is a solution to the given equation, whereas $x=0$ is not.

Answer: $\frac{y^2}{2}+y+\ln |y-1|=-\frac{1}{x}+C,$ $y=1.$

Equations reducible to equations with separable variables.

Equations of the form $y^{\prime }=f(ax+by)$ are reduced to equations with separable variables by the substitution $z=ax+by$ (or $z=ax+by+c,$ where $c$ is any constant).

Examples.

1. $y^{\prime }=\sqrt{4x+2y-1}.$

Solution.

Let's make the substitution.

$z=4x+2y-1.$ From this, $y=\frac{1}{2}(z-4x+1)\Rightarrow y^{\prime }=\frac{1}{2}{z}^{\prime }-2$

We obtain$$\frac{1}{2}{z}^{\prime }-2=\sqrt{z}$$

Let's transform this equation into the form (2).

$$\frac{dz}{2dx}=\sqrt{z}+2\Rightarrow \frac{dz}{\sqrt{z}+2}=2dx$$

We integrate both sides of the equation:

$$\int \frac{dz}{\sqrt{z}+2}=\int 2dx$$

$$\int \frac{dz}{\sqrt{z}+2}=[z=t^{2}dz=2tdt]=\int \frac{2tdt}{t+2}=2\int \frac{t+2-2}{t+2}dt=2(t-2\ln |t+2|)+C=$$ $$=2(\sqrt{z}-2\ln |\sqrt{z}+2|)+C=2(\sqrt{4x+2y-1}-2\ln |\sqrt{4x+2y-1}+2|)+C.$$

$$\int 2dx=2x+C.$$

Thus, we obtained:

$$2(\sqrt{4x+2y-1}-2\ln |\sqrt{4x+2y-1}+2|)=2x+C\Rightarrow$$ $$\Rightarrow \sqrt{4x+2y-1}-\ln (\sqrt{4x+2y-1}+2)=x+C.$$

Answer: $$\sqrt{4x+2y-1}-\ln (\sqrt{4x+2y-1}+2)=x+C..$$

Tags: differential equations, equations with separable variables