Question

Y'/x -2y=(1-x²)e^x^2 дифференциальное уравнение вас

Answer 1

$\frac{y'}{x}-2y=(1-x^2)\cdot e^{x^2}\; |\cdot x\ne o\\\\y'-2xy=(1-x^2)\cdot x\cdot e^{x^2}\\\\y=uv\; ,\; \; y'=u'v+uv'\\\\u'v+uv'-2uvx=(1-x^2)\cdot x\cdot e^{x^2}\\\\u'v+u\cdot (v'-2vx)=(1-x^2)\cdot x\cdot e^{x^2}\\\\a)\; \; \frac{dv}{dx}-2vx=0\; ,\; \; \int \frac{dv}{v}=\int 2x\, dx \; ,\; \; ln|v|=2\cdot \frac{x^2}{2}\\\\v=e^{x^2}\\\\b)\; \; u'\cdot e^{x^2}=(1-x^2)\cdot x\cdot e^{x^2}\\\\\int du=\int (1-x^2)\cdot x\cdot dx\\\\\int du=\int (x-x^3)\cdot dx\\\\u=\frac{x^2}{2}-\frac{x^4}{4}+C\\\\c)\; \; y=e^{x^2}\cdot (\frac{x^2}{2}-\frac{x^4}{4}+C)$