Question

Решить задачу коши пожалуйста срочно ​

Answer 1

Первое уравнение

$2x^2\frac{\mathrm{d} y}{\mathrm{d} x}y+y^2=2\Leftrightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-y^2+2}{2x^2y}\Leftrightarrow 2\frac{1}{2-y^2}\cdot y\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{1}{x^2}\Rightarrow \\\Rightarrow \int 2\frac{1}{2-y^2}\cdot y\frac{\mathrm{d} y}{\mathrm{d} x}dx=\int \frac{dx}{x^2}\Rightarrow \ln\left ( y^2-2 \right )=-\frac{1}{x}+c_1\Rightarrow$$\\\Rightarrow y=\pm \sqrt{e^{1/x+c_1}+2}\overset{y(1)=0}{\Rightarrow }\sqrt{c_1e+2}=0\Rightarrow c_1=-\frac{2}{e}\Rightarrow y=\pm\sqrt{-2e^{1/x-1}+2}$

Второе уравнение

$\mu =\exp\left \{ \int \mathrm{tg}2x dx \right \}=\frac{1}{\sqrt{\cos 2x}}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+y\mathrm{tg}2x=\sin 2x\Leftrightarrow \\\Leftrightarrow \frac{1}{\sqrt{\cos 2x}}\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{y\sin 2x}{\cos^{3/2}2x}=\frac{\sin 2x}{\sqrt{\cos 2x}}\Leftrightarrow$$\Leftrightarrow \frac{1}{\sqrt{\cos 2x}}+\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{\sqrt{\cos 2x}} \right )y=\frac{\sin 2x}{\sqrt{\cso 2x}}\Leftrightarrow \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{\sqrt{\cos 2x}} \right )=\frac{\sin 2x}{\sqrt{\cos 2x}}\Rightarrow$$\Rightarrow \int \frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{y}{\sqrt{\cos 2x}} \right )dx=\int \frac{\sin 2x}{\sqrt{\cos 2x}}dx\Rightarrow \frac{y}{\sqrt{\cos 2x}}=-\sqrt{\cos 2x}+c_1\Rightarrow \\\Rightarrow y=-\cos 2x+c_1\sqrt{\cos 2x}$