Czy ktoś może wyjaśnić gdzie popełniam błąd?
\(x^2y'-y=2\)
\(x^2 \frac{dy}{dx} -y=2\)
\(x^2 \frac{dy}{dx} =y\)
\(x^2dy=ydx/y\)
\(\frac{x^2dy}{y}=dx/x^2\)
\(\frac{dy}{y}= \frac{dx}{x^2}\)
\(\int \frac{dy}{y} = \int \frac{dx}{x^2}\)
\(ln y = \int \frac{1}{x^2}dx\)
\(\int \frac{1}{x^2}dx= \int x^{-2}=- \frac{1}{x}+c\)
\(ln y = - \frac{1}{x}+c\)
\(y = Ce^{-\frac{1}{x}}\)
\(y = C(x)e^{-\frac{1}{x}}\)
\(y' = [C(x)e^{-\frac{1}{x}}]'=C'(x)e^{-\frac{1}{x}}+C(x)e^{-\frac{1}{x}}\)
\(x^2[C'(x)e^{-\frac{1}{x}}+C(x)e^{-\frac{1}{x}}]-C(x)e^{-\frac{1}{x}}=2\)
Oblicz równanie.
Otrzymałeś(aś) rozwiązanie do zamieszczonego zadania? - podziękuj autorowi rozwiązania! Kliknij
\([C'(x)e^{- \frac{1}{x}}]'=C'(x)e^{- \frac{1}{x}}+C(x)(e^{- \frac{1}{x}})'=C'(x)e^{- \frac{1}{x}}+C(x)e^{- \frac{1}{x}} \cdot (- \frac{1}{x^2})=C'(x)e^{- \frac{1}{x}}- \frac{C(x)e^{- \frac{1}{x}}}{x^2}\)
\(x^2C'(x)e^{- \frac{1}{x}}- \frac{C(x)e^{- \frac{1}{x}}}{x^2}]+C(x)e^{ -\frac{1}{x} }=2\)
\(x^2C'(x)e^{- \frac{1}{x}}=2/ \cdot e^{ \frac{1}{x}}\)
\(x^2C'(x)=2e^{ \frac{1}{x}} /x^2\)
\(C'(x)= \frac{2e^{ \frac{1}{x}}}{x^2}\)
\(C(x)= \int\frac{2e^{ \frac{1}{x}}}{x^2}dx\)
\(\int \frac{2e^{ \frac{1}{x}}}{x^2}dx =\)
\(u=x^2, v'=2e^ \frac{1}{x}\)
\(u'=2x, v= 2e^ \frac{1}{x}+c\)
\(x^2 \cdot 2e^{ \frac{1}{x} }- \int 2x \cdot 2e^ \frac{1}{x}=x^2 \cdot 2e^{ \frac{1}{x} }- 4\int xe^ \frac{1}{x}\)
\(u=x, v'=e^ \frac{1}{x}\)
\(u'=1, v= e^ \frac{1}{x} +c\)
\(xe^ \frac{1}{x}- \int e^ \frac{1}{x}dx =xe^ \frac{1}{x}- e^ \frac{1}{x}\)
\(C(x)=xe^ \frac{1}{x}- e^ \frac{1}{x}\)
\((xe^ \frac{1}{x}- e^ \frac{1}{x})e^ {-\frac{1}{x}} =2\)
\(x^2C'(x)e^{- \frac{1}{x}}- \frac{C(x)e^{- \frac{1}{x}}}{x^2}]+C(x)e^{ -\frac{1}{x} }=2\)
\(x^2C'(x)e^{- \frac{1}{x}}=2/ \cdot e^{ \frac{1}{x}}\)
\(x^2C'(x)=2e^{ \frac{1}{x}} /x^2\)
\(C'(x)= \frac{2e^{ \frac{1}{x}}}{x^2}\)
\(C(x)= \int\frac{2e^{ \frac{1}{x}}}{x^2}dx\)
\(\int \frac{2e^{ \frac{1}{x}}}{x^2}dx =\)
\(u=x^2, v'=2e^ \frac{1}{x}\)
\(u'=2x, v= 2e^ \frac{1}{x}+c\)
\(x^2 \cdot 2e^{ \frac{1}{x} }- \int 2x \cdot 2e^ \frac{1}{x}=x^2 \cdot 2e^{ \frac{1}{x} }- 4\int xe^ \frac{1}{x}\)
\(u=x, v'=e^ \frac{1}{x}\)
\(u'=1, v= e^ \frac{1}{x} +c\)
\(xe^ \frac{1}{x}- \int e^ \frac{1}{x}dx =xe^ \frac{1}{x}- e^ \frac{1}{x}\)
\(C(x)=xe^ \frac{1}{x}- e^ \frac{1}{x}\)
\((xe^ \frac{1}{x}- e^ \frac{1}{x})e^ {-\frac{1}{x}} =2\)