MCQ
વિકલ સમીકરણ $\log_e\left(\frac{dy}{dx}\right)=3x+4y,y(0)=0$ નો વિશિષ્ટ ઉકેલ $4e^{3x}+3e^{-4y}=\ .....$ છે.
- A$6$
- B$4$
- ✓$7$
- D$3$
✓
Answer
Correct option: C.
$7$
$\log_e\left(\frac{dy}{dx}\right)=3x+4y$
$\therefore\frac{dy}{dx}=e^{3x+4y}=e^{3x}\cdot e^{4y}$
$\therefore\frac{dy}{e^{4y}}=e^{3x}dx$
$\therefore\int e^{-4y}dy=\int e^{3x}dx+c$
$\therefore\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+c$
$\therefore-3e^{-4y}=4e^{3x}-12c\ \ \ \ \ \ \ \ .....(1)$
હવે, $x=0$ ત્યારે $y=0$
$\therefore-3e^0=4e^0-12c$
$\therefore12c=4+3=7$
$\therefore$ સમીકરણ $(1)$ પરથી, $-3e^{-4y}=4e^{3x}-7$
$\therefore4e^{3x}+3e^{-4y}=7$
$\therefore\frac{dy}{dx}=e^{3x+4y}=e^{3x}\cdot e^{4y}$
$\therefore\frac{dy}{e^{4y}}=e^{3x}dx$
$\therefore\int e^{-4y}dy=\int e^{3x}dx+c$
$\therefore\frac{e^{-4y}}{-4}=\frac{e^{3x}}{3}+c$
$\therefore-3e^{-4y}=4e^{3x}-12c\ \ \ \ \ \ \ \ .....(1)$
હવે, $x=0$ ત્યારે $y=0$
$\therefore-3e^0=4e^0-12c$
$\therefore12c=4+3=7$
$\therefore$ સમીકરણ $(1)$ પરથી, $-3e^{-4y}=4e^{3x}-7$
$\therefore4e^{3x}+3e^{-4y}=7$
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