Question

Срочно!!!!!!!!! Решить 2 дифференциальных уравнений​

Answer 1

1.

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

$2) \ x\cdot u'v=-x$

$x\cdot \frac{du}{dx}\cdot x^2 =-x \\\\ du = -\frac{x}{x^3} \, dx \\ \\ du =-x^{-2} \, dx \\ \\ \int {1} \, du =- \int {x^{-2}} \, dx \\ \\ u=-(-x^{-1})+C =\frac{1}{x}+C \\ \\ y=uv=(\frac{1}{x}+C)\cdot x^2=x+Cx^2$

2.

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

$2) \ x^2\cdot u'v=-3 \\ \\ x^2\cdot \frac{du}{dx}\cdot x^2 =-3 \\ \\ du=-\frac{3}{x^4} \\ \\ \int {1} \, du =-3\int {x^{-4}} \, dx \\ \\ u=(-3)\cdot \frac{x^{-3}}{(-3)}+C \\ \\ u=\frac{1}{x^3}+C \\ \\ y=uv=(\frac{1}{x^3}+C)\cdot x^2 = \frac{1}{x}+Cx^2$