Question

Найти частное решение(частный интеграл) дифференциального уравнения.
$(1-x^2)y'+xy=1, y(0)=1$

Answer 1

$(1-x^2)\, y'+xy=1\; \; ,\; \; y())=1\\\\y'+\frac{x}{1-x^2}\cdot y=\frac{1}{1-x^2}\\\\y=uv\; ,\; y'=u'v+uv'\\\\u'v+uv'+\frac{x}{1-x^2}\cdot uv=\frac{1}{1-x^2}\\\\u'v+u\cdot (v'+\frac{x}{1-x^2}\cdot v)=\frac{1}{1-x^2}\\\\a)\; \; v'+\frac{x}{1-x^2}\cdot v=0\; \; ,\; \; \int \frac{dv}{v}=\int \frac{-x\, dx}{1-x^2}\\\\ln|v|=\frac{1}{2}\cdot ln|1-x^2|\; \; ,\; \; v=\sqrt{1-x^2}\\\\b)\; \; u'v=\frac{1}{1-x^2}\; \; ,\; \; u'\cdot \sqrt{1-x^2}=\frac{1}{1-x^2}\; ,\\\\\int du=\int \frac{dx}{\sqrt{(1-x^2)^3}}$

$\int \frac{dx}{\sqrt{(1-x^2)^3}}=\Big [\; x=sint\; ,\; dx=cost\, dt\; ,\; 1-x^2=1-sin^2t=cos^2t\; \Big ]=\\\\=\int \frac{cost\, dt}{\sqrt{cos^6t}}=\int \frac{cost\,dt}{cos^3t}=\int \frac{dt}{cos^2t}=tgt+C_1=tg(arcsinx)+C_1=\frac{x}{\sqrt{1-x^2}}+C_1\; ;\\\\u=\frac{x}{\sqrt{1-x^2}}+C\; ;\\\\c)\; \; y=uv\; ,\; \; y=\sqrt{1-x^2}\cdot (\frac{x}{\sqrt{1-x^2}} +C)=x+C\sqrt{1-x^2}\; ;\\\\d)\; \; y(0)=1:\; \; 1=0+C\cdot \sqrt{1-0^2}\; ,\; \; C=1\\\\y_{chastn.}=x+\sqrt{1-x^2}$