Solve $\dfrac{dy}{dx} = e^{x+y}$.
A$e^{-y} = -e^x + c$
B$e^y = e^x + c$
C$y = e^x + c$
D$\log y = x + c$
Answer & Solution
Correct answer: A. $e^{-y} = -e^x + c$
$dy/e^y = e^x\,dx$ ⇒ $\int e^{-y}\,dy = \int e^x\,dx$ ⇒ $-e^{-y} = e^x + c'$ ⇒ $e^{-y} = -e^x + c$.
Related questions
Reducing $(2x-1)\dfrac{d^2y}{dx^2}-2\dfrac{dy}{dx}=0$ with $y=2$ and $\dfrac{dy}{dx}=3$ atA village grows at a rate proportional to its population. It was 20,000 in 1999 and 25,000For $\dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}+2y=6e^{x}+ in x$, the particular integral is:The complementary function of $\dfrac{d^2y}{dx^2}+3\dfrac{dy}{dx}+2y=6e^{x}+ in x$ is:Solutions of a differential equation form a set that is a:The power counted for the degree must be a whole number that is:An equation whose highest derivative is the third has order:The study at this stage is confined to equations that are: