Question

Найти общее решение дифференциального уравнения.

Answer 1

Ответ:

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

$\displaystyle b)\ \ u'\cdot \frac{1}{1+x^2}=\frac{2x^2}{1+x^2}\ \ ,\ \ \ \int du=\int 2x^2\, dx\ \ ,\ \ \ u=\frac{2x^3}{3}+C_1\ \ ,\\\\\\c)\ \ t=\frac{1}{1+x^2}\cdot \Big(\frac{2x^3}{3}+C_1\Big)\\\\\\d)\ \ y'=\frac{1}{1+x^2}\cdot \Big(\frac{2x^3}{3}+C_1\Big)\ \ ,\ \ \ y'=\frac{2x^3}{3\, (1+x^2)}+\frac{C_1}{1+x^2}\ \ ,\\\\\\\int dy=\frac{2}{3}\int \frac{x^3\, dx}{1+x^2}+\int \frac{C_1\, dx}{1+x^2}\\\\\\y=\frac{2}{3}\int \Big(x-\frac{x}{1+x^2}\Big)\, dx+C_1\int \frac{dx}{1+x^2}$

$\displaystyle y=\frac{2}{3}\cdot \Big(\frac{x^2}{2}-\frac{1}{2}\cdot ln(1+x^2)\Big)+C_1\, arctgx+C_2$