Đáp án:
$6)\quad A.\ y = e^{\tfrac{\pi}{4} - \arctan x}$
$7)\quad C.\ 2e^{\tfrac{y^2}{2}}= \sqrt e(1 + e^x)$
Giải thích các bước giải:
$6)\quad (1+x^2)dy + ydx = 0$
$\Leftrightarrow \dfrac1ydy = - \dfrac{1}{1+x^2}dx$
$\Leftrightarrow \ln y = - \arctan x + C$
Ta có: $y(1)= 1$
$\Leftrightarrow \ln 1 = - \arctan 1 + C$
$\Leftrightarrow C = \dfrac{\pi}{4}$
Do đó:
$\quad \ln y = - \arctan x + \dfrac{\pi}{4}$
$\Leftrightarrow y = e^{\tfrac{\pi}{4} - \arctan x}$
$7)\quad (1+e^x)yy' = e^x$
$\Leftrightarrow ydy = \dfrac{e^x}{1 + e^x}dx$
$\Leftrightarrow \dfrac{y^2}{2} = \ln(1+e^x) + C$
Ta có:
$\quad y(0)= -1$
$\Leftrightarrow \dfrac{(-1)^2}{2}= \ln(1+e^0) + C$
$\Leftrightarrow C = \dfrac12 - \ln2$
Do đó:
$\quad \dfrac{y^2}{2} = \ln(1+e^x) + \dfrac12 - \ln2$
$\Leftrightarrow e^{\tfrac{y^2}{2}}= e^{\ln(1+e^x) + \tfrac12 - \ln2}$
$\Leftrightarrow e^{\tfrac{y^2}{2}}= e^{\ln(1+e^x)}.e^{\tfrac12 - \ln2}$
$\Leftrightarrow e^{\tfrac{y^2}{2}}= (1+e^x)\cdot \dfrac{e^{\tfrac12}}{e^{\ln2}}$
$\Leftrightarrow e^{\tfrac{y^2}{2}}= (1+e^x)\cdot \dfrac{\sqrt e}{2}$
$\Leftrightarrow 2e^{\tfrac{y^2}{2}}= \sqrt e(1+e^x)$
