Question

Помогите решить задачу Коши!

Answer 1

Ответ:

$\pm 4.12$

Пошаговое объяснение:

$(x+3)ydy+(y^{2}+1)dx=0;$

$(x+3)ydy=-(y^{2}+1)dx;$

$\frac{(x+3)ydy}{dx}=-(y^{2}+1);$

$\frac{x+3}{dx}=\frac{-(y^{2}+1)}{ydy};$

$\frac{dx}{x+3}=\frac{ydy}{-(y^{2}+1)};$

$\int\ {\frac{d(x+3)}{x+3}} \ = -\int\ {\frac{y}{y^{2}+1}} \, dy ;$

$d(y^{2}+1)=2ydy;$

$ln |x+3|=-\frac{1}{2} \int\ {\frac{2ydy}{y^{2}+1}} \ \quad | \quad \cdot (-1)$

$-ln |x+3|=\frac{1}{2} \int\ {\frac{d(y^{2}+1)}{y^{2}+1}} \ ;$

$ln |\frac{1}{x+3}|=\frac{1}{2}(ln(y^{2}+1)+ ln |C|);$

$ln |\frac{1}{x+3}|=\frac{1}{2} ln |C(y^{2}+1)|;$

$ln |\frac{1}{x+3}|=ln |\sqrt{C(y^{2}+1)}|;$

$\sqrt{C(y^{2}+1)}=\frac{1}{x+3};$

$C(y^{2}+1)=\frac{1}{(x+3)^{2}};$

$y^{2}+1=\frac{1}{C(x+3)^{2}};$

$y^{2}=\frac{1}{C(x+3)^{2}}-1;$

$y= \pm \sqrt{\frac{1}{C(x+3)^{2}}-1};$

$y(0)=1 \Rightarrow \pm \sqrt{\frac{1}{C(0+3)^{2}}-1}=1 \Rightarrow \frac{1}{9C}-1=1 \Rightarrow \frac{1}{9C}=2 \Rightarrow 9C=\frac{1}{2} \Rightarrow C=\frac{1}{18};$

$y= \pm \sqrt{\frac{1}{\frac{1}{18}(x+3)^{2}}-1};$

$y= \pm \sqrt{\frac{18}{(x+3)^{2}}-1};$

$y(-2)= \pm \sqrt{\frac{18}{(-2+3)^{2}}-1}= \pm \sqrt{\frac{18}{1^{2}}-1}= \pm \sqrt{18-1}= \pm \sqrt{17};$

$\pm \sqrt{17} \approx \pm 4.12;$