Question

Найти общее решение дифференциального уравнения.

Answer 1

Ответ:

$\displaystyle (1-x^2)\, y''-xy'=2\\\\y''-\frac{x}{1-x^2}\, y'=\frac{2}{1-x^2}\ \ \ ,\ \ \ y'=t(x)\ \ ,\ \ y''=t'(x)\\\\\\t'-\frac{x}{1-x^2}\cdot t=\frac{2}{1-x^2}\ \ ,\ \ \ \ \ t=uv\ ,\ t'=u'v+uv'\\\\\\u'v+uv'-\frac{x}{1-x^2}\cdot uv=\frac{2}{1-x^2}\ \ ,\ \ \ u'v+u\, (\, v'-\frac{x}{1-x^2}\, v\ )=\frac{2}{1-x^2}\ \ ,\\\\\\a)\ \ v'-\frac{x}{1-x^2}\, v=0\ \ , \ \ \ \int \frac{dv}{v}=\int \frac{x\, dx}{1-x^2}\ \ ,\ \ ln\, v=-\frac{1}{2}\, ln(1-x^2)\ ,\\\\\\v=\frac{1}{\sqrt{1-x^2}}$

$\displaystyle b)\ \ u'v=\frac{2}{1-x^2}\ \ ,\ \ \ \frac{du}{dx}\cdot \frac{1}{\sqrt{1-t=x^2}}=\frac{2}{1-x^2}\ \ ,\ \ \int du=\int \frac{2\, dx}{\sqrt{1-x^2}}\ \ ,\\\\\\u=2\, arcsinx+C_1\\\\\\c)\ \ t=\frac{1}{\sqrt{1-x^2}}\cdot (2arcsinx+C_1)\ \ ,\ \ y'=\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}\cdot (2arcsinx+C_1)\ ,\\\\\\\int dy=\int \frac{2\, arcsinx+C_1}{\sqrt{1-x^2}}\, dx\ \ ,\ \ y=\int \frac{2\, arcsinx}{\sqrt{1-x^2}}\, dx+\int \frac{C_1}{\sqrt{1-x^2}}\, dx\\\\\\\\\boxed{\ y=arcsin^2x+C_1\, arcsinx+C_2\ }$