Question

Помогите с дифференциальными уравнениями
Помогите я ща помру, я не понимаю, я дурак

Answer 1

$2x+2xy^2+\sqrt{2-x^2} y'=0 \\ \\ \sqrt{2-x^2} y'=-2x\cdot (1+y^2) \\ \\ \frac{y'}{1+y^2}=-\frac{2x}{\sqrt{2-x^2}} \\ \\ \int {\frac{dy}{1+y^2}} = - \int\ {\frac{2x}{\sqrt{2-x^2}}} \, dx \\ \\ t = (2-x^2); \ \ dt = -2x \, dx \\ 2xdx=dv; \\ \\ arctg \, y= \int {\frac{dt}{\sqrt{t}}} \\ \\ arctg \, y= \int {t^{-\frac{1}{2}}} \, dt \\ \\ arctg \, y=2\cdot \sqrt{t}+C \\ \\ arctg \, y=2 \cdot \sqrt{2-x^2}+C \\ \\ y=tg \, (2\sqrt{2-x^2}+C)$

$xy'-y+xe^{\frac{y}{x}}=0; \\ \\ y=x\cdot v; \ \ \ y'=x\cdot v'+v \\ \\ x\cdot ( x\cdot v'+v)-x\cdot v+x\cdot e^{v}=0; \\ \\ x\cdot (x\cdot v'+v-v+e^{v})=0; \\ \\ x\cdot (x\cdot v'+e^v)=0 \\ \\ x^2\cdot v'+x\cdot e^v=0 \\ \\ x^2\cdot v'=-x\cdot e^v \\ \\ \frac{v'}{e^v}=-\frac{1}{x} \\ \\ \int {\frac{dv}{e^v}}=-\int {\frac{dx}{x}} \\ \\ \int {e^{-v} \, dv} = -\int {\frac{dx}{x}} \\ \\ -e^{-v}=-\ln{x}+C \\ \\ e^{-v}=\ln {x} +C \\ \\ -v=\ln{(\ln{x}+C)} \\ \\ v=-\ln{(\ln{x}+C)}$

$\frac{y}{x}=-\ln{(\ln{x}+C)} \\ \\ y= -x\cdot \ln{(\ln{x}+C)}$

$y'+tg \, x\cdot y= \frac{1}{\cos{x}} \\ \\ y=uv; \ \ \ y=u'v+uv' \\ \\ u'v+uv'+tg \, x\cdot uv=\frac{1}{\cos{x}} \\ \\ u'v+u\cdot (v'+tg \, x\cdot v)=\frac{1}{\cos{x}} \\ \\ \left \{ {{v'+tg \, x \cdot v =0} \atop {u'v=\frac{1}{\cos{x}}}} \right. \\ \\ v'+tg \, x \cdot v =0 \\ \\ \frac{v'}{v}=-tg \, x \\ \\ \int {\frac{dv}{v}}=-\int {\frac{\sin{x}}{\cos{x}}} \, dx; \ \ \ \ \ \ \ \ \ \ d(\cos{x})=-\sin{x}dx\\ \\ \ln{v}=-(-\int {\frac{d(\cos{x})}{\cos{x}}}) \\ \\ \ln {v}=\ln{(\cos{x}})$

$v=\cos{x} \\\\ u'\cdot \cos{x}=\frac{1}{\cos{x}}\\ \\ \frac{du}{dx}=\frac{1}{\cos^2{x}} \\\\ \ u=\int{\frac{dx}{\cos^2{x}}} = tg \, x+C \\ \\ y=uv=tg \, x +C \cdot \cos{x}=\sin{x}+C\cdot\cos{x}$

$y''\cdot tg \,x-y'-1=0\\ \\ y'=v; \ \ y''=v' \\ \\ v'\cdot tg \, x-v-1=0\\ \\ v' tg \, x =v+1 \\ \\ \frac{v'}{v+1}=\frac{1}{tg \, x } \\ \\ \int \frac{dv}{v+1}=\int {\frac{\cos{x} }{\sin{x} } \, dx } ; \ \ \ \ \ \ \ d(\sin{x})=\cos{x}dx\\ \\ \ln{(v+1)}=\ln{(C_1\sin{x})} \\ \\ v+1=C_1\sin{x}; \ \ \ \ v =C_1\sin{x}-1 \\ \\ y'=C_1\sin{x}-1 \\ \\ y=\int {(C_1\sin{x}-1)} \, dx =-C_1\cos{x}-x+C_2$