Question

Помогите решить дифференциальное уравнение.
Y’=2^x+y

Answer 1

Ответ:

$y'=2^{x}+y\\\\y'-y=2^{x}\\\\y=uv\ ,\ y'=u'v+uv'\\\\u'v+uv'-uv=2^{x}\\\\u'v+u\, (v'-v)=2^{x}\\\\a)\ \ v'-v=0\ \ ,\ \ \dfrac{dv}{dx}=v\ \ ,\ \ \ \int \dfrac{dv}{v}=\int dx\ \ ,\ \ \ lnv=x\ \ ,\ \ v=e^{x}\\\\\\b)\ \ u'\cdot e^{x}=2^{x}\ \ ,\ \ \ \dfrac{du}{dx}=\dfrac{2^{x}}{e^{x}}\ \ ,\ \ \ \int du=\int \Big(\dfrac{2}{e}\Big)^{x}\, dx\ \ ,\\\\u=\dfrac{\Big(\dfrac{2}{e}\Big)^{x}}{ln\dfrac{2}{e}}+C\ \ ,\ \ \ u=\dfrac{2^{x}}{e^{x}\cdot (ln2-1)}+C$

$c)\ \ y=e^{x}\cdot \Big(\dfrac{2^{x}}{e^{x}\cdot (ln2-1)}+C\Big)\ \ \ ,\ \ \ \ y=\dfrac{2^{x}}{ln2-1}+C\cdot e^{x}$