What is the solution for xy^'+(x+1/3)y=e^{-2x}y^{-2} ?

The solution for xy^'+(x+1/3)y=e^{-2x}y^{-2} is y=\sqrt[3]{(3/(e^{2x)}+(c_{1})/(e^{3x)})/x}

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xy^'+(x+1/3)y=e^{-2x}y^{-2}

x y^{^{'}} + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

y = 3 \frac{\frac{3}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}}{x}

Solution steps

x y^{'} (x) + (x + \frac{1}{3}) y = e^{- 2 x} y^{- 2}

Solve Bernoulli ODE:

y = 3 \frac{\frac{3}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}}{x}

y = 3 \frac{\frac{3}{e ^{2 x}} + \frac{c _{1}}{e ^{3 x}}}{x}

t an g e n t f (x) = x^{2} - x + 4, at x = 1

ma c l a u r in \frac{1}{3 x + 8}

t an g e n t f (x) = x^{2} - 4, (4, 12)

\int \frac{4 x + 6}{( x ^{2} + 2 ) ( x + 2 )} d x

(x + y) d x + (x + 2 y) d y = 0

What is the solution for xy^'+(x+1/3)y=e^{-2x}y^{-2} ?
The solution for xy^'+(x+1/3)y=e^{-2x}y^{-2} is y=\sqrt[3]{(3/(e^{2x)}+(c_{1})/(e^{3x)})/x}