Question

Решить дифференциальные уравнения:
1) y'=x²*eˣ
2) (1+y)dx-(1-x)dy=0
3) (1+x²)dy-2x(y+3)dx=0, y(0)=-1

Answer 1

Ответ:

$1)\ \ y'=x^2\cdot e^{x}\ \ ,\ \ \ \dfrac{dy}{dx}=x^2\cdot e^{x}\ \ ,\ \ \ \ \int dy=\int x^2\cdot e^{x}\, dx\ \ \ ,\\\\\int x^2\cdot e^{x}\, dx=\Big[\ u=x^2\ ,\ du=2x\, dx\ ,\ dv=e^{x}\, dx\ ,\ v=e^{x}\ \Big]=\\\\=x^2\cdot e^{x}-2\int x\cdot e^{x}\, dx=\Big[\ u=x\ ,\ du=dx\ ,\ v=e^{x}\ \Big]=\\\\=x^2\cdot e^{x}-2\cdot \Big(x\cdot e^{x}-\int e^{x}\, dx\Big)=x^2\cdot e^{x}-2x\cdot e^{x}+2\, e^{x}+C=\\\\=e^{x}\cdot (x^2-2x+2)+C$

$2)\ \ (1+y)\, dx-(1-x)\, dy=0\\\\\int \dfrac{dx}{1-x}=\int \dfrac{dy}{1+y}\\\\-ln|1-x|=ln|1+y|+ln|C|\\\\\dfrac{1}{1-x}=C\cdot (1+y)\ \ \ ,\ \ \ 1+y=\dfrac{1}{C\, (1-x)}\ \ ,\ \ \ y=\dfrac{1}{C\, (1-x)}-1\ .$

$3)\ \ (1+x^2)\, dy-2x\, (y+3)\, dx=0\ \ ,\ \ \ y(0)=1\\\\\int \dfrac{dy}{y+3}=\int \dfrac{2x\, dx}{x^2+1} \ \ ,\ \ \ ln|y+3|=ln|x^2+1|+ln|C|\ \ \ ,\\\\\\y+3=C\, (x^2+1)\ \ ,\ \ \ y=C\, (x^2+1)-3\\\\y(0)=1:\ \ 1=C\, (0+1)-3\ \ ,\ \ C=4\\\\y_{chastn.}=4\, (x^2+1)-3$