Question

Розв'язати диференціальне рівняння: фото $y'+y\sqrt{x} =1+\sqrt{x}$

!!!!Без неповної гамма функції!!!!

Answer 1

$\mu =\exp\left \{ \int \sqrt{x}dx \right \}=\exp\left \{ \frac{2x^{3/2}}{3} \right \}\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}+\sqrt{x}y=\sqrt{x}+1\Leftrightarrow\\\Leftrightarrow \exp\left \{ \frac{2x^{3/2}}{3} \right \}\frac{\mathrm{d} y}{\mathrm{d} x}+\left ( \exp\left \{ \frac{2x^{3/2}}{3}\sqrt{x} \right \} \right )y=-\exp\left \{ \frac{2x^{3/2}}{3} \right \}\left ( -\sqrt{x}-1 \right )\Leftrightarrow$$\Leftrightarrow \frac{\mathrm{d} }{\mathrm{d} x}\left ( \exp\left \{ \frac{2x^{3/2}}{3}y \right \} \right )=-\exp\left \{ \frac{2x^{3/2}}{3} \right \}\left ( -\sqrt{x}-1 \right )\Leftrightarrow \\\Leftrightarrow \int \frac{\mathrm{d} }{\mathrm{d} x}\left ( \exp\left \{ \frac{2x^{3/2}}{3} \right \}y \right )dx=-\int \exp\left \{ \frac{2x^{3/2}}{3} \right \}\left ( -\sqrt{x}-1 \right )dx\Leftrightarrow$$\Leftrightarrow \exp\left \{ \frac{2x^{3/2}}{3} \right \}y=\exp\left \{ \frac{2x^{3/2}}{3} \right \}-\frac{1}{\left ( -x^{3/2} \right )^{2/3}}\sqrt[3]{\frac{2}{3}}\Gamma \left ( \frac{2}{3},-\frac{2x^{3/2}}{3} \right )+C$